Related
I'm trying to solve a problem which goes like this:
Problem
Given an array of integers "arr" of size "n", process two types of queries. There are "q" queries you need to answer.
Query type 1
input: l r
result: output number of inversions in [l, r]
Query type 2
input: x y
result: update the value at arr [x] to y
Inversion
For every index j < i, if arr [j] > arr [i], the pair (j, i) is one inversion.
Input
n = 5
q = 3
arr = {1, 4, 3, 5, 2}
queries:
type = 1, l = 1, r = 5
type = 2, x = 1, y = 4
type = 1, l = 1, r = 5
Output
4
6
Constraints
Time: 4 secs
1 <= n, q <= 100000
1 <= arr [i] <= 40
1 <= l, r, x <= n
1 <= y <= 40
I know how to solve a simpler version of this problem without updates, i.e. to simply count the number of inversions for each position using a segment tree or fenwick tree in O(N*log(N)). The only solution I have to this problem is O(q*N*log(N)) (I think) with segment tree other than the O(q*N2) trivial algorithm. This however does not fit within the time constraints of the problem. I would like to have hints towards a better algorithm to solve the problem in O(N*log(N)) (if it's possible) or maybe O(N*log2(N)).
I first came across this problem two days ago and have been spending a few hours here and there to try and solve it. However, I'm finding it non-trivial to do so and would like to have some help/hints regarding the same. Thanks for your time and patience.
Updates
Solution
With the suggestion, answer and help by Tanvir Wahid, I've implemented the source code for the problem in C++ and would like to share it here for anyone who might stumble across this problem and not have an intuitive idea on how to solve it. Thank you!
Let's build a segment tree with each node containing information about how many inversions exist and the frequency count of elements present in its segment of authority.
node {
integer inversion_count : 0
array [40] frequency : {0...0}
}
Building the segment tree and handling updates
For each leaf node, initialise inversion count to 0 and increase frequency of the represented element from the input array to 1. The frequency of the parent nodes can be calculated by summing up frequencies of the left and right childrens. The inversion count of parent nodes can be calculated by summing up the inversion counts of left and right children nodes added with the new inversions created upon merging the two segments of their authority which can be calculated using the frequencies of elements in each child. This calculation basically finds out the product of frequencies of bigger elements in the left child and frequencies of smaller elements in the right child.
parent.inversion_count = left.inversion_count + right.inversion_count
for i in [39, 0]
for j in [0, i)
parent.inversion_count += left.frequency [i] * right.frequency [j]
Updates are handled similarly.
Answering range queries on inversion counts
To answer the query for the number of inversions in the range [l, r], we calculate the inversions using the source code attached below.
Time Complexity: O(q*log(n))
Note
The source code attached does break some good programming habits. The sole purpose of the code is to "solve" the given problem and not to accomplish anything else.
Source Code
/**
Lost Arrow (Aryan V S)
Saturday 2020-10-10
**/
#include "bits/stdc++.h"
using namespace std;
struct node {
int64_t inv = 0;
vector <int> freq = vector <int> (40, 0);
void combine (const node& l, const node& r) {
inv = l.inv + r.inv;
for (int i = 39; i >= 0; --i) {
for (int j = 0; j < i; ++j) {
// frequency of bigger numbers in the left * frequency of smaller numbers on the right
inv += 1LL * l.freq [i] * r.freq [j];
}
freq [i] = l.freq [i] + r.freq [i];
}
}
};
void build (vector <node>& tree, vector <int>& a, int v, int tl, int tr) {
if (tl == tr) {
tree [v].inv = 0;
tree [v].freq [a [tl]] = 1;
}
else {
int tm = (tl + tr) / 2;
build(tree, a, 2 * v + 1, tl, tm);
build(tree, a, 2 * v + 2, tm + 1, tr);
tree [v].combine(tree [2 * v + 1], tree [2 * v + 2]);
}
}
void update (vector <node>& tree, int v, int tl, int tr, int pos, int val) {
if (tl == tr) {
tree [v].inv = 0;
tree [v].freq = vector <int> (40, 0);
tree [v].freq [val] = 1;
}
else {
int tm = (tl + tr) / 2;
if (pos <= tm)
update(tree, 2 * v + 1, tl, tm, pos, val);
else
update(tree, 2 * v + 2, tm + 1, tr, pos, val);
tree [v].combine(tree [2 * v + 1], tree [2 * v + 2]);
}
}
node inv_cnt (vector <node>& tree, int v, int tl, int tr, int l, int r) {
if (l > r)
return node();
if (tl == l && tr == r)
return tree [v];
int tm = (tl + tr) / 2;
node result;
result.combine(inv_cnt(tree, 2 * v + 1, tl, tm, l, min(r, tm)), inv_cnt(tree, 2 * v + 2, tm + 1, tr, max(l, tm + 1), r));
return result;
}
void solve () {
int n, q;
cin >> n >> q;
vector <int> a (n);
for (int i = 0; i < n; ++i) {
cin >> a [i];
--a [i];
}
vector <node> tree (4 * n);
build(tree, a, 0, 0, n - 1);
while (q--) {
int type, x, y;
cin >> type >> x >> y;
--x; --y;
if (type == 1) {
node result = inv_cnt(tree, 0, 0, n - 1, x, y);
cout << result.inv << '\n';
}
else if (type == 2) {
update(tree, 0, 0, n - 1, x, y);
}
else
assert(false);
}
}
int main () {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
std::cout.precision(10);
std::cout << std::fixed << std::boolalpha;
int t = 1;
// std::cin >> t;
while (t--)
solve();
return 0;
}
arr[i] can be at most 40. We can use this to our advantage. What we need is a segment tree. Each node will hold 41 values (A long long int which represents inversions for this range and a array of size 40 for count of each numbers. A struct will do). How do we merge two children of a node. We know inversions for left child and right child. Also know frequency of each numbers in both of them. Inversion of parent node will be summation of inversions of both children plus number of inversions between left and right child. We can easily find inversions between two children from frequency of numbers. Query can be done in similar way. Complexity O(40*qlog(n))
I came to this problem in a challenge.
There are two arrays A and B both of size of N and we need to return the count of pairs (A[i],B[j]) where gcd(A[i],B[j])==1 and A[i] != B[j].
I could only think of brute force approach which exceeded time limit for few test cases.
for(int i=0; i<n; i++) {
for(int j=0; j<n; j++) {
if(__gcd(a[i],b[j])==1) {
printf("%d %d\n", a[i], b[j]);
}
}
}
Can you advice time efficient algorithm to solve this.
Edit: Not able to share question link as this was from a hiring challenge. Adding the constraints and input/output format as I remember.
Input -
First line will contain N, the number of elements present in both arrays.
Second line will contain N space separated integers, elements of array A.
Third line will contain N space separated integers, elements of array B.
Output -
The count of pairs A[i],A[j] as per the conditions.
Constraints -
1 <= N <= 10^5
1 < A[i],B[j] <= 10^9 where i,j < N
The first step is to use Eratosthenes sieve to calculate the prime numbers up to sqrt(10^9). This sieve can then be used to quickly find all prime factors of any number less than 10^9 (see the getPrimeFactors(...) function in the code sample below).
Next, for each A[i] with prime factors p0, p1, ..., pk, we compute all possible sub-products X - p0, p1, p0p1, p2, p0p2, p1p2, p0p1p2, p3, p0p3, ..., p0p1p2...pk and count them in map cntp[X]. Effectively, the map cntp[X] tells us the number of elements A[i] divisible by X, where X is a product of prime numbers to the power of 0 or 1. So for example, for the number A[i] = 12, the prime factors are 2, 3. We will count cntp[2]++, cntp[3]++ and cntp[6]++.
Finally, for each B[j] with prime factors p0, p1, ..., pk, we again compute all possible sub-products X and use the Inclusion-exclusion principle to count all non-coprime pairs C_j (i.e. the number of A[i]s that share at least one prime factor with B[j]). The numbers C_j are then subtracted from the total number of pairs - N*N to get the final answer.
Note: the Inclusion-exclusion principle looks like this:
C_j = (cntp[p0] + cntp[p1] + ... + cntp[pk]) -
(cntp[p0p1] + cntp[p0p2] + ... + cntp[pk-1pk]) +
(cntp[p0p1p2] + cntp[p0p1p3] + ... + cntp[pk-2pk-1pk]) -
...
and accounts for the fact that in cntp[X] and cntp[Y] we could have counted the same number A[i] twice, given that it is divisible by both X and Y.
Here is a possible C++ implementation of the algorithm, which produces the same results as the naive O(n^2) algorithm by OP:
// get prime factors of a using pre-generated sieve
std::vector<int> getPrimeFactors(int a, const std::vector<int> & primes) {
std::vector<int> f;
for (auto p : primes) {
if (p > a) break;
if (a % p == 0) {
f.push_back(p);
do {
a /= p;
} while (a % p == 0);
}
}
if (a > 1) f.push_back(a);
return f;
}
// find coprime pairs A_i and B_j
// A_i and B_i <= 1e9
void solution(const std::vector<int> & A, const std::vector<int> & B) {
// generate prime sieve
std::vector<int> primes;
primes.push_back(2);
for (int i = 3; i*i <= 1e9; ++i) {
bool isPrime = true;
for (auto p : primes) {
if (i % p == 0) {
isPrime = false;
break;
}
}
if (isPrime) {
primes.push_back(i);
}
}
int N = A.size();
struct Entry {
int n = 0;
int64_t p = 0;
};
// cntp[X] - number of times the product X can be expressed
// with prime factors of A_i
std::map<int64_t, int64_t> cntp;
for (int i = 0; i < N; i++) {
auto f = getPrimeFactors(A[i], primes);
// count possible products using non-repeating prime factors of A_i
std::vector<Entry> x;
x.push_back({ 0, 1 });
for (auto p : f) {
int k = x.size();
for (int i = 0; i < k; ++i) {
int nn = x[i].n + 1;
int64_t pp = x[i].p*p;
++cntp[pp];
x.push_back({ nn, pp });
}
}
}
// use Inclusion–exclusion principle to count non-coprime pairs
// and subtract them from the total number of prairs N*N
int64_t cnt = N; cnt *= N;
for (int i = 0; i < N; i++) {
auto f = getPrimeFactors(B[i], primes);
std::vector<Entry> x;
x.push_back({ 0, 1 });
for (auto p : f) {
int k = x.size();
for (int i = 0; i < k; ++i) {
int nn = x[i].n + 1;
int64_t pp = x[i].p*p;
x.push_back({ nn, pp });
if (nn % 2 == 1) {
cnt -= cntp[pp];
} else {
cnt += cntp[pp];
}
}
}
}
printf("cnt = %d\n", (int) cnt);
}
Live example
I cannot estimate the complexity analytically, but here are some profiling result on my laptop for different N and uniformly random A[i] and B[j]:
For N = 1e2, takes ~0.02 sec
For N = 1e3, takes ~0.05 sec
For N = 1e4, takes ~0.38 sec
For N = 1e5, takes ~3.80 sec
For comparison, the O(n^2) approach takes:
For N = 1e2, takes ~0.00 sec
For N = 1e3, takes ~0.15 sec
For N = 1e4, takes ~15.1 sec
For N = 1e5, takes too long, didn't wait to finish
Python Implementation:
import math
from collections import defaultdict
def sieve(MAXN):
spf = [0 for i in range(MAXN)]
spf[1] = 1
for i in range(2, MAXN):
spf[i] = i
for i in range(4, MAXN, 2):
spf[i] = 2
for i in range(3, math.ceil(math.sqrt(MAXN))):
if (spf[i] == i):
for j in range(i * i, MAXN, i):
if (spf[j] == j):
spf[j] = i
return(spf)
def getFactorization(x,spf):
ret = list()
while (x != 1):
ret.append(spf[x])
x = x // spf[x]
return(list(set(ret)))
def coprime_pairs(N,A,B):
MAXN=max(max(A),max(B))+1
spf=sieve(MAXN)
cntp=defaultdict(int)
for i in range(N):
f=getFactorization(A[i],spf)
x=[[0,1]]
for p in f:
k=len(x)
for i in range(k):
nn=x[i][0]+1
pp=x[i][1]*p
cntp[pp]+=1
x.append([nn,pp])
cnt=0
for i in range(N):
f=getFactorization(B[i],spf)
x=[[0,1]]
for p in f:
k=len(x)
for i in range(k):
nn=x[i][0]+1
pp=x[i][1]*p
x.append([nn,pp])
if(nn%2==1):
cnt+=cntp[pp]
else:
cnt-=cntp[pp]
return(N*N-cnt)
import random
N=10001
A=[random.randint(1,N) for _ in range(N)]
B=[random.randint(1,N) for _ in range(N)]
print(coprime_pairs(N,A,B))
I want to write a function that takes an array of letters as an argument and a number of those letters to select.
Say you provide an array of 8 letters and want to select 3 letters from that. Then you should get:
8! / ((8 - 3)! * 3!) = 56
Arrays (or words) in return consisting of 3 letters each.
Art of Computer Programming Volume 4: Fascicle 3 has a ton of these that might fit your particular situation better than how I describe.
Gray Codes
An issue that you will come across is of course memory and pretty quickly, you'll have problems by 20 elements in your set -- 20C3 = 1140. And if you want to iterate over the set it's best to use a modified gray code algorithm so you aren't holding all of them in memory. These generate the next combination from the previous and avoid repetitions. There are many of these for different uses. Do we want to maximize the differences between successive combinations? minimize? et cetera.
Some of the original papers describing gray codes:
Some Hamilton Paths and a Minimal Change Algorithm
Adjacent Interchange Combination Generation Algorithm
Here are some other papers covering the topic:
An Efficient Implementation of the Eades, Hickey, Read Adjacent Interchange Combination Generation Algorithm (PDF, with code in Pascal)
Combination Generators
Survey of Combinatorial Gray Codes (PostScript)
An Algorithm for Gray Codes
Chase's Twiddle (algorithm)
Phillip J Chase, `Algorithm 382: Combinations of M out of N Objects' (1970)
The algorithm in C...
Index of Combinations in Lexicographical Order (Buckles Algorithm 515)
You can also reference a combination by its index (in lexicographical order). Realizing that the index should be some amount of change from right to left based on the index we can construct something that should recover a combination.
So, we have a set {1,2,3,4,5,6}... and we want three elements. Let's say {1,2,3} we can say that the difference between the elements is one and in order and minimal. {1,2,4} has one change and is lexicographically number 2. So the number of 'changes' in the last place accounts for one change in the lexicographical ordering. The second place, with one change {1,3,4} has one change but accounts for more change since it's in the second place (proportional to the number of elements in the original set).
The method I've described is a deconstruction, as it seems, from set to the index, we need to do the reverse – which is much trickier. This is how Buckles solves the problem. I wrote some C to compute them, with minor changes – I used the index of the sets rather than a number range to represent the set, so we are always working from 0...n.
Note:
Since combinations are unordered, {1,3,2} = {1,2,3} --we order them to be lexicographical.
This method has an implicit 0 to start the set for the first difference.
Index of Combinations in Lexicographical Order (McCaffrey)
There is another way:, its concept is easier to grasp and program but it's without the optimizations of Buckles. Fortunately, it also does not produce duplicate combinations:
The set that maximizes , where .
For an example: 27 = C(6,4) + C(5,3) + C(2,2) + C(1,1). So, the 27th lexicographical combination of four things is: {1,2,5,6}, those are the indexes of whatever set you want to look at. Example below (OCaml), requires choose function, left to reader:
(* this will find the [x] combination of a [set] list when taking [k] elements *)
let combination_maccaffery set k x =
(* maximize function -- maximize a that is aCb *)
(* return largest c where c < i and choose(c,i) <= z *)
let rec maximize a b x =
if (choose a b ) <= x then a else maximize (a-1) b x
in
let rec iterate n x i = match i with
| 0 -> []
| i ->
let max = maximize n i x in
max :: iterate n (x - (choose max i)) (i-1)
in
if x < 0 then failwith "errors" else
let idxs = iterate (List.length set) x k in
List.map (List.nth set) (List.sort (-) idxs)
A small and simple combinations iterator
The following two algorithms are provided for didactic purposes. They implement an iterator and (a more general) folder overall combinations.
They are as fast as possible, having the complexity O(nCk). The memory consumption is bound by k.
We will start with the iterator, which will call a user provided function for each combination
let iter_combs n k f =
let rec iter v s j =
if j = k then f v
else for i = s to n - 1 do iter (i::v) (i+1) (j+1) done in
iter [] 0 0
A more general version will call the user provided function along with the state variable, starting from the initial state. Since we need to pass the state between different states we won't use the for-loop, but instead, use recursion,
let fold_combs n k f x =
let rec loop i s c x =
if i < n then
loop (i+1) s c ##
let c = i::c and s = s + 1 and i = i + 1 in
if s < k then loop i s c x else f c x
else x in
loop 0 0 [] x
In C#:
public static IEnumerable<IEnumerable<T>> Combinations<T>(this IEnumerable<T> elements, int k)
{
return k == 0 ? new[] { new T[0] } :
elements.SelectMany((e, i) =>
elements.Skip(i + 1).Combinations(k - 1).Select(c => (new[] {e}).Concat(c)));
}
Usage:
var result = Combinations(new[] { 1, 2, 3, 4, 5 }, 3);
Result:
123
124
125
134
135
145
234
235
245
345
Short java solution:
import java.util.Arrays;
public class Combination {
public static void main(String[] args){
String[] arr = {"A","B","C","D","E","F"};
combinations2(arr, 3, 0, new String[3]);
}
static void combinations2(String[] arr, int len, int startPosition, String[] result){
if (len == 0){
System.out.println(Arrays.toString(result));
return;
}
for (int i = startPosition; i <= arr.length-len; i++){
result[result.length - len] = arr[i];
combinations2(arr, len-1, i+1, result);
}
}
}
Result will be
[A, B, C]
[A, B, D]
[A, B, E]
[A, B, F]
[A, C, D]
[A, C, E]
[A, C, F]
[A, D, E]
[A, D, F]
[A, E, F]
[B, C, D]
[B, C, E]
[B, C, F]
[B, D, E]
[B, D, F]
[B, E, F]
[C, D, E]
[C, D, F]
[C, E, F]
[D, E, F]
May I present my recursive Python solution to this problem?
def choose_iter(elements, length):
for i in xrange(len(elements)):
if length == 1:
yield (elements[i],)
else:
for next in choose_iter(elements[i+1:], length-1):
yield (elements[i],) + next
def choose(l, k):
return list(choose_iter(l, k))
Example usage:
>>> len(list(choose_iter("abcdefgh",3)))
56
I like it for its simplicity.
Lets say your array of letters looks like this: "ABCDEFGH". You have three indices (i, j, k) indicating which letters you are going to use for the current word, You start with:
A B C D E F G H
^ ^ ^
i j k
First you vary k, so the next step looks like that:
A B C D E F G H
^ ^ ^
i j k
If you reached the end you go on and vary j and then k again.
A B C D E F G H
^ ^ ^
i j k
A B C D E F G H
^ ^ ^
i j k
Once you j reached G you start also to vary i.
A B C D E F G H
^ ^ ^
i j k
A B C D E F G H
^ ^ ^
i j k
...
Written in code this look something like that
void print_combinations(const char *string)
{
int i, j, k;
int len = strlen(string);
for (i = 0; i < len - 2; i++)
{
for (j = i + 1; j < len - 1; j++)
{
for (k = j + 1; k < len; k++)
printf("%c%c%c\n", string[i], string[j], string[k]);
}
}
}
The following recursive algorithm picks all of the k-element combinations from an ordered set:
choose the first element i of your combination
combine i with each of the combinations of k-1 elements chosen recursively from the set of elements larger than i.
Iterate the above for each i in the set.
It is essential that you pick the rest of the elements as larger than i, to avoid repetition. This way [3,5] will be picked only once, as [3] combined with [5], instead of twice (the condition eliminates [5] + [3]). Without this condition you get variations instead of combinations.
Short example in Python:
def comb(sofar, rest, n):
if n == 0:
print sofar
else:
for i in range(len(rest)):
comb(sofar + rest[i], rest[i+1:], n-1)
>>> comb("", "abcde", 3)
abc
abd
abe
acd
ace
ade
bcd
bce
bde
cde
For explanation, the recursive method is described with the following example:
Example: A B C D E
All combinations of 3 would be:
A with all combinations of 2 from the rest (B C D E)
B with all combinations of 2 from the rest (C D E)
C with all combinations of 2 from the rest (D E)
I found this thread useful and thought I would add a Javascript solution that you can pop into Firebug. Depending on your JS engine, it could take a little time if the starting string is large.
function string_recurse(active, rest) {
if (rest.length == 0) {
console.log(active);
} else {
string_recurse(active + rest.charAt(0), rest.substring(1, rest.length));
string_recurse(active, rest.substring(1, rest.length));
}
}
string_recurse("", "abc");
The output should be as follows:
abc
ab
ac
a
bc
b
c
In C++ the following routine will produce all combinations of length distance(first,k) between the range [first,last):
#include <algorithm>
template <typename Iterator>
bool next_combination(const Iterator first, Iterator k, const Iterator last)
{
/* Credits: Mark Nelson http://marknelson.us */
if ((first == last) || (first == k) || (last == k))
return false;
Iterator i1 = first;
Iterator i2 = last;
++i1;
if (last == i1)
return false;
i1 = last;
--i1;
i1 = k;
--i2;
while (first != i1)
{
if (*--i1 < *i2)
{
Iterator j = k;
while (!(*i1 < *j)) ++j;
std::iter_swap(i1,j);
++i1;
++j;
i2 = k;
std::rotate(i1,j,last);
while (last != j)
{
++j;
++i2;
}
std::rotate(k,i2,last);
return true;
}
}
std::rotate(first,k,last);
return false;
}
It can be used like this:
#include <string>
#include <iostream>
int main()
{
std::string s = "12345";
std::size_t comb_size = 3;
do
{
std::cout << std::string(s.begin(), s.begin() + comb_size) << std::endl;
} while (next_combination(s.begin(), s.begin() + comb_size, s.end()));
return 0;
}
This will print the following:
123
124
125
134
135
145
234
235
245
345
static IEnumerable<string> Combinations(List<string> characters, int length)
{
for (int i = 0; i < characters.Count; i++)
{
// only want 1 character, just return this one
if (length == 1)
yield return characters[i];
// want more than one character, return this one plus all combinations one shorter
// only use characters after the current one for the rest of the combinations
else
foreach (string next in Combinations(characters.GetRange(i + 1, characters.Count - (i + 1)), length - 1))
yield return characters[i] + next;
}
}
Simple recursive algorithm in Haskell
import Data.List
combinations 0 lst = [[]]
combinations n lst = do
(x:xs) <- tails lst
rest <- combinations (n-1) xs
return $ x : rest
We first define the special case, i.e. selecting zero elements. It produces a single result, which is an empty list (i.e. a list that contains an empty list).
For n > 0, x goes through every element of the list and xs is every element after x.
rest picks n - 1 elements from xs using a recursive call to combinations. The final result of the function is a list where each element is x : rest (i.e. a list which has x as head and rest as tail) for every different value of x and rest.
> combinations 3 "abcde"
["abc","abd","abe","acd","ace","ade","bcd","bce","bde","cde"]
And of course, since Haskell is lazy, the list is gradually generated as needed, so you can partially evaluate exponentially large combinations.
> let c = combinations 8 "abcdefghijklmnopqrstuvwxyz"
> take 10 c
["abcdefgh","abcdefgi","abcdefgj","abcdefgk","abcdefgl","abcdefgm","abcdefgn",
"abcdefgo","abcdefgp","abcdefgq"]
And here comes granddaddy COBOL, the much maligned language.
Let's assume an array of 34 elements of 8 bytes each (purely arbitrary selection.) The idea is to enumerate all possible 4-element combinations and load them into an array.
We use 4 indices, one each for each position in the group of 4
The array is processed like this:
idx1 = 1
idx2 = 2
idx3 = 3
idx4 = 4
We vary idx4 from 4 to the end. For each idx4 we get a unique combination
of groups of four. When idx4 comes to the end of the array, we increment idx3 by 1 and set idx4 to idx3+1. Then we run idx4 to the end again. We proceed in this manner, augmenting idx3,idx2, and idx1 respectively until the position of idx1 is less than 4 from the end of the array. That finishes the algorithm.
1 --- pos.1
2 --- pos 2
3 --- pos 3
4 --- pos 4
5
6
7
etc.
First iterations:
1234
1235
1236
1237
1245
1246
1247
1256
1257
1267
etc.
A COBOL example:
01 DATA_ARAY.
05 FILLER PIC X(8) VALUE "VALUE_01".
05 FILLER PIC X(8) VALUE "VALUE_02".
etc.
01 ARAY_DATA OCCURS 34.
05 ARAY_ITEM PIC X(8).
01 OUTPUT_ARAY OCCURS 50000 PIC X(32).
01 MAX_NUM PIC 99 COMP VALUE 34.
01 INDEXXES COMP.
05 IDX1 PIC 99.
05 IDX2 PIC 99.
05 IDX3 PIC 99.
05 IDX4 PIC 99.
05 OUT_IDX PIC 9(9).
01 WHERE_TO_STOP_SEARCH PIC 99 COMP.
* Stop the search when IDX1 is on the third last array element:
COMPUTE WHERE_TO_STOP_SEARCH = MAX_VALUE - 3
MOVE 1 TO IDX1
PERFORM UNTIL IDX1 > WHERE_TO_STOP_SEARCH
COMPUTE IDX2 = IDX1 + 1
PERFORM UNTIL IDX2 > MAX_NUM
COMPUTE IDX3 = IDX2 + 1
PERFORM UNTIL IDX3 > MAX_NUM
COMPUTE IDX4 = IDX3 + 1
PERFORM UNTIL IDX4 > MAX_NUM
ADD 1 TO OUT_IDX
STRING ARAY_ITEM(IDX1)
ARAY_ITEM(IDX2)
ARAY_ITEM(IDX3)
ARAY_ITEM(IDX4)
INTO OUTPUT_ARAY(OUT_IDX)
ADD 1 TO IDX4
END-PERFORM
ADD 1 TO IDX3
END-PERFORM
ADD 1 TO IDX2
END_PERFORM
ADD 1 TO IDX1
END-PERFORM.
Another C# version with lazy generation of the combination indices. This version maintains a single array of indices to define a mapping between the list of all values and the values for the current combination, i.e. constantly uses O(k) additional space during the entire runtime. The code generates individual combinations, including the first one, in O(k) time.
public static IEnumerable<T[]> Combinations<T>(this T[] values, int k)
{
if (k < 0 || values.Length < k)
yield break; // invalid parameters, no combinations possible
// generate the initial combination indices
var combIndices = new int[k];
for (var i = 0; i < k; i++)
{
combIndices[i] = i;
}
while (true)
{
// return next combination
var combination = new T[k];
for (var i = 0; i < k; i++)
{
combination[i] = values[combIndices[i]];
}
yield return combination;
// find first index to update
var indexToUpdate = k - 1;
while (indexToUpdate >= 0 && combIndices[indexToUpdate] >= values.Length - k + indexToUpdate)
{
indexToUpdate--;
}
if (indexToUpdate < 0)
yield break; // done
// update combination indices
for (var combIndex = combIndices[indexToUpdate] + 1; indexToUpdate < k; indexToUpdate++, combIndex++)
{
combIndices[indexToUpdate] = combIndex;
}
}
}
Test code:
foreach (var combination in new[] {'a', 'b', 'c', 'd', 'e'}.Combinations(3))
{
System.Console.WriteLine(String.Join(" ", combination));
}
Output:
a b c
a b d
a b e
a c d
a c e
a d e
b c d
b c e
b d e
c d e
Here is an elegant, generic implementation in Scala, as described on 99 Scala Problems.
object P26 {
def flatMapSublists[A,B](ls: List[A])(f: (List[A]) => List[B]): List[B] =
ls match {
case Nil => Nil
case sublist#(_ :: tail) => f(sublist) ::: flatMapSublists(tail)(f)
}
def combinations[A](n: Int, ls: List[A]): List[List[A]] =
if (n == 0) List(Nil)
else flatMapSublists(ls) { sl =>
combinations(n - 1, sl.tail) map {sl.head :: _}
}
}
If you can use SQL syntax - say, if you're using LINQ to access fields of an structure or array, or directly accessing a database that has a table called "Alphabet" with just one char field "Letter", you can adapt following code:
SELECT A.Letter, B.Letter, C.Letter
FROM Alphabet AS A, Alphabet AS B, Alphabet AS C
WHERE A.Letter<>B.Letter AND A.Letter<>C.Letter AND B.Letter<>C.Letter
AND A.Letter<B.Letter AND B.Letter<C.Letter
This will return all combinations of 3 letters, notwithstanding how many letters you have in table "Alphabet" (it can be 3, 8, 10, 27, etc.).
If what you want is all permutations, rather than combinations (i.e. you want "ACB" and "ABC" to count as different, rather than appear just once) just delete the last line (the AND one) and it's done.
Post-Edit: After re-reading the question, I realise what's needed is the general algorithm, not just a specific one for the case of selecting 3 items. Adam Hughes' answer is the complete one, unfortunately I cannot vote it up (yet). This answer's simple but works only for when you want exactly 3 items.
I had a permutation algorithm I used for project euler, in python:
def missing(miss,src):
"Returns the list of items in src not present in miss"
return [i for i in src if i not in miss]
def permutation_gen(n,l):
"Generates all the permutations of n items of the l list"
for i in l:
if n<=1: yield [i]
r = [i]
for j in permutation_gen(n-1,missing([i],l)): yield r+j
If
n<len(l)
you should have all combination you need without repetition, do you need it?
It is a generator, so you use it in something like this:
for comb in permutation_gen(3,list("ABCDEFGH")):
print comb
https://gist.github.com/3118596
There is an implementation for JavaScript. It has functions to get k-combinations and all combinations of an array of any objects. Examples:
k_combinations([1,2,3], 2)
-> [[1,2], [1,3], [2,3]]
combinations([1,2,3])
-> [[1],[2],[3],[1,2],[1,3],[2,3],[1,2,3]]
Lets say your array of letters looks like this: "ABCDEFGH". You have three indices (i, j, k) indicating which letters you are going to use for the current word, You start with:
A B C D E F G H
^ ^ ^
i j k
First you vary k, so the next step looks like that:
A B C D E F G H
^ ^ ^
i j k
If you reached the end you go on and vary j and then k again.
A B C D E F G H
^ ^ ^
i j k
A B C D E F G H
^ ^ ^
i j k
Once you j reached G you start also to vary i.
A B C D E F G H
^ ^ ^
i j k
A B C D E F G H
^ ^ ^
i j k
...
function initializePointers($cnt) {
$pointers = [];
for($i=0; $i<$cnt; $i++) {
$pointers[] = $i;
}
return $pointers;
}
function incrementPointers(&$pointers, &$arrLength) {
for($i=0; $i<count($pointers); $i++) {
$currentPointerIndex = count($pointers) - $i - 1;
$currentPointer = $pointers[$currentPointerIndex];
if($currentPointer < $arrLength - $i - 1) {
++$pointers[$currentPointerIndex];
for($j=1; ($currentPointerIndex+$j)<count($pointers); $j++) {
$pointers[$currentPointerIndex+$j] = $pointers[$currentPointerIndex]+$j;
}
return true;
}
}
return false;
}
function getDataByPointers(&$arr, &$pointers) {
$data = [];
for($i=0; $i<count($pointers); $i++) {
$data[] = $arr[$pointers[$i]];
}
return $data;
}
function getCombinations($arr, $cnt)
{
$len = count($arr);
$result = [];
$pointers = initializePointers($cnt);
do {
$result[] = getDataByPointers($arr, $pointers);
} while(incrementPointers($pointers, count($arr)));
return $result;
}
$result = getCombinations([0, 1, 2, 3, 4, 5], 3);
print_r($result);
Based on https://stackoverflow.com/a/127898/2628125, but more abstract, for any size of pointers.
Here you have a lazy evaluated version of that algorithm coded in C#:
static bool nextCombination(int[] num, int n, int k)
{
bool finished, changed;
changed = finished = false;
if (k > 0)
{
for (int i = k - 1; !finished && !changed; i--)
{
if (num[i] < (n - 1) - (k - 1) + i)
{
num[i]++;
if (i < k - 1)
{
for (int j = i + 1; j < k; j++)
{
num[j] = num[j - 1] + 1;
}
}
changed = true;
}
finished = (i == 0);
}
}
return changed;
}
static IEnumerable Combinations<T>(IEnumerable<T> elements, int k)
{
T[] elem = elements.ToArray();
int size = elem.Length;
if (k <= size)
{
int[] numbers = new int[k];
for (int i = 0; i < k; i++)
{
numbers[i] = i;
}
do
{
yield return numbers.Select(n => elem[n]);
}
while (nextCombination(numbers, size, k));
}
}
And test part:
static void Main(string[] args)
{
int k = 3;
var t = new[] { "dog", "cat", "mouse", "zebra"};
foreach (IEnumerable<string> i in Combinations(t, k))
{
Console.WriteLine(string.Join(",", i));
}
}
Hope this help you!
Another version, that forces all the first k to appear firstly, then all the first k+1 combinations, then all the first k+2 etc.. It means that if you have sorted array, the most important on the top, it would take them and expand gradually to the next ones - only when it is must do so.
private static bool NextCombinationFirstsAlwaysFirst(int[] num, int n, int k)
{
if (k > 1 && NextCombinationFirstsAlwaysFirst(num, num[k - 1], k - 1))
return true;
if (num[k - 1] + 1 == n)
return false;
++num[k - 1];
for (int i = 0; i < k - 1; ++i)
num[i] = i;
return true;
}
For instance, if you run the first method ("nextCombination") on k=3, n=5 you'll get:
0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4
But if you'll run
int[] nums = new int[k];
for (int i = 0; i < k; ++i)
nums[i] = i;
do
{
Console.WriteLine(string.Join(" ", nums));
}
while (NextCombinationFirstsAlwaysFirst(nums, n, k));
You'll get this (I added empty lines for clarity):
0 1 2
0 1 3
0 2 3
1 2 3
0 1 4
0 2 4
1 2 4
0 3 4
1 3 4
2 3 4
It's adding "4" only when must to, and also after "4" was added it adds "3" again only when it must to (after doing 01, 02, 12).
Array.prototype.combs = function(num) {
var str = this,
length = str.length,
of = Math.pow(2, length) - 1,
out, combinations = [];
while(of) {
out = [];
for(var i = 0, y; i < length; i++) {
y = (1 << i);
if(y & of && (y !== of))
out.push(str[i]);
}
if (out.length >= num) {
combinations.push(out);
}
of--;
}
return combinations;
}
Clojure version:
(defn comb [k l]
(if (= 1 k) (map vector l)
(apply concat
(map-indexed
#(map (fn [x] (conj x %2))
(comb (dec k) (drop (inc %1) l)))
l))))
Algorithm:
Count from 1 to 2^n.
Convert each digit to its binary representation.
Translate each 'on' bit to elements of your set, based on position.
In C#:
void Main()
{
var set = new [] {"A", "B", "C", "D" }; //, "E", "F", "G", "H", "I", "J" };
var kElement = 2;
for(var i = 1; i < Math.Pow(2, set.Length); i++) {
var result = Convert.ToString(i, 2).PadLeft(set.Length, '0');
var cnt = Regex.Matches(Regex.Escape(result), "1").Count;
if (cnt == kElement) {
for(int j = 0; j < set.Length; j++)
if ( Char.GetNumericValue(result[j]) == 1)
Console.Write(set[j]);
Console.WriteLine();
}
}
}
Why does it work?
There is a bijection between the subsets of an n-element set and n-bit sequences.
That means we can figure out how many subsets there are by counting sequences.
e.g., the four element set below can be represented by {0,1} X {0, 1} X {0, 1} X {0, 1} (or 2^4) different sequences.
So - all we have to do is count from 1 to 2^n to find all the combinations. (We ignore the empty set.) Next, translate the digits to their binary representation. Then substitute elements of your set for 'on' bits.
If you want only k element results, only print when k bits are 'on'.
(If you want all subsets instead of k length subsets, remove the cnt/kElement part.)
(For proof, see MIT free courseware Mathematics for Computer Science, Lehman et al, section 11.2.2. https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/readings/ )
short python code, yielding index positions
def yield_combos(n,k):
# n is set size, k is combo size
i = 0
a = [0]*k
while i > -1:
for j in range(i+1, k):
a[j] = a[j-1]+1
i=j
yield a
while a[i] == i + n - k:
i -= 1
a[i] += 1
All said and and done here comes the O'caml code for that.
Algorithm is evident from the code..
let combi n lst =
let rec comb l c =
if( List.length c = n) then [c] else
match l with
[] -> []
| (h::t) -> (combi t (h::c))#(combi t c)
in
combi lst []
;;
Here is a method which gives you all combinations of specified size from a random length string. Similar to quinmars' solution, but works for varied input and k.
The code can be changed to wrap around, ie 'dab' from input 'abcd' w k=3.
public void run(String data, int howMany){
choose(data, howMany, new StringBuffer(), 0);
}
//n choose k
private void choose(String data, int k, StringBuffer result, int startIndex){
if (result.length()==k){
System.out.println(result.toString());
return;
}
for (int i=startIndex; i<data.length(); i++){
result.append(data.charAt(i));
choose(data,k,result, i+1);
result.setLength(result.length()-1);
}
}
Output for "abcde":
abc abd abe acd ace ade bcd bce bde cde
Short javascript version (ES 5)
let combine = (list, n) =>
n == 0 ?
[[]] :
list.flatMap((e, i) =>
combine(
list.slice(i + 1),
n - 1
).map(c => [e].concat(c))
);
let res = combine([1,2,3,4], 3);
res.forEach(e => console.log(e.join()));
Another python recusive solution.
def combination_indicies(n, k, j = 0, stack = []):
if len(stack) == k:
yield list(stack)
return
for i in range(j, n):
stack.append(i)
for x in combination_indicies(n, k, i + 1, stack):
yield x
stack.pop()
list(combination_indicies(5, 3))
Output:
[[0, 1, 2],
[0, 1, 3],
[0, 1, 4],
[0, 2, 3],
[0, 2, 4],
[0, 3, 4],
[1, 2, 3],
[1, 2, 4],
[1, 3, 4],
[2, 3, 4]]
I created a solution in SQL Server 2005 for this, and posted it on my website: http://www.jessemclain.com/downloads/code/sql/fn_GetMChooseNCombos.sql.htm
Here is an example to show usage:
SELECT * FROM dbo.fn_GetMChooseNCombos('ABCD', 2, '')
results:
Word
----
AB
AC
AD
BC
BD
CD
(6 row(s) affected)
Here is my proposition in C++
I tried to impose as little restriction on the iterator type as i could so this solution assumes just forward iterator, and it can be a const_iterator. This should work with any standard container. In cases where arguments don't make sense it throws std::invalid_argumnent
#include <vector>
#include <stdexcept>
template <typename Fci> // Fci - forward const iterator
std::vector<std::vector<Fci> >
enumerate_combinations(Fci begin, Fci end, unsigned int combination_size)
{
if(begin == end && combination_size > 0u)
throw std::invalid_argument("empty set and positive combination size!");
std::vector<std::vector<Fci> > result; // empty set of combinations
if(combination_size == 0u) return result; // there is exactly one combination of
// size 0 - emty set
std::vector<Fci> current_combination;
current_combination.reserve(combination_size + 1u); // I reserve one aditional slot
// in my vector to store
// the end sentinel there.
// The code is cleaner thanks to that
for(unsigned int i = 0u; i < combination_size && begin != end; ++i, ++begin)
{
current_combination.push_back(begin); // Construction of the first combination
}
// Since I assume the itarators support only incrementing, I have to iterate over
// the set to get its size, which is expensive. Here I had to itrate anyway to
// produce the first cobination, so I use the loop to also check the size.
if(current_combination.size() < combination_size)
throw std::invalid_argument("combination size > set size!");
result.push_back(current_combination); // Store the first combination in the results set
current_combination.push_back(end); // Here I add mentioned earlier sentinel to
// simplyfy rest of the code. If I did it
// earlier, previous statement would get ugly.
while(true)
{
unsigned int i = combination_size;
Fci tmp; // Thanks to the sentinel I can find first
do // iterator to change, simply by scaning
{ // from right to left and looking for the
tmp = current_combination[--i]; // first "bubble". The fact, that it's
++tmp; // a forward iterator makes it ugly but I
} // can't help it.
while(i > 0u && tmp == current_combination[i + 1u]);
// Here is probably my most obfuscated expression.
// Loop above looks for a "bubble". If there is no "bubble", that means, that
// current_combination is the last combination, Expression in the if statement
// below evaluates to true and the function exits returning result.
// If the "bubble" is found however, the ststement below has a sideeffect of
// incrementing the first iterator to the left of the "bubble".
if(++current_combination[i] == current_combination[i + 1u])
return result;
// Rest of the code sets posiotons of the rest of the iterstors
// (if there are any), that are to the right of the incremented one,
// to form next combination
while(++i < combination_size)
{
current_combination[i] = current_combination[i - 1u];
++current_combination[i];
}
// Below is the ugly side of using the sentinel. Well it had to haave some
// disadvantage. Try without it.
result.push_back(std::vector<Fci>(current_combination.begin(),
current_combination.end() - 1));
}
}
Here is a code I recently wrote in Java, which calculates and returns all the combination of "num" elements from "outOf" elements.
// author: Sourabh Bhat (heySourabh#gmail.com)
public class Testing
{
public static void main(String[] args)
{
// Test case num = 5, outOf = 8.
int num = 5;
int outOf = 8;
int[][] combinations = getCombinations(num, outOf);
for (int i = 0; i < combinations.length; i++)
{
for (int j = 0; j < combinations[i].length; j++)
{
System.out.print(combinations[i][j] + " ");
}
System.out.println();
}
}
private static int[][] getCombinations(int num, int outOf)
{
int possibilities = get_nCr(outOf, num);
int[][] combinations = new int[possibilities][num];
int arrayPointer = 0;
int[] counter = new int[num];
for (int i = 0; i < num; i++)
{
counter[i] = i;
}
breakLoop: while (true)
{
// Initializing part
for (int i = 1; i < num; i++)
{
if (counter[i] >= outOf - (num - 1 - i))
counter[i] = counter[i - 1] + 1;
}
// Testing part
for (int i = 0; i < num; i++)
{
if (counter[i] < outOf)
{
continue;
} else
{
break breakLoop;
}
}
// Innermost part
combinations[arrayPointer] = counter.clone();
arrayPointer++;
// Incrementing part
counter[num - 1]++;
for (int i = num - 1; i >= 1; i--)
{
if (counter[i] >= outOf - (num - 1 - i))
counter[i - 1]++;
}
}
return combinations;
}
private static int get_nCr(int n, int r)
{
if(r > n)
{
throw new ArithmeticException("r is greater then n");
}
long numerator = 1;
long denominator = 1;
for (int i = n; i >= r + 1; i--)
{
numerator *= i;
}
for (int i = 2; i <= n - r; i++)
{
denominator *= i;
}
return (int) (numerator / denominator);
}
}
im reading about base b expansion of n algorithm and this book is really hard to read and understand, i am not sure what the bottom part means...
does it return n or what ? how would you do this...thanks
some method (n,b)
if b == 0
return 1
q = n
k = 0
while q does not == 0
a[k] = q % b
q = q / b
k = k + 1
return ???
I wrote an implementation in C for the function. It uses a pointer as input parameter, where the output (a vector of integers) will be placed. The function also returns an integer - the logical size of the vector.
#include <assert.h>
int toBase(int n, int b, int* answer) {
assert(b > 1);
q = n
k = 0
while (q != 0) {
answer[k] = q % b;
q /= b;
++k;
}
return k;
}
int main() {
int answer[32];
int n = 100000;
int b = 2;
int answerSize = toBase(n, b, answer);
// use answer and answerSize
return 0;
}
Another way to do it (without the pointer parameter) is to allocate memory for the vector inside the function and return it, requiring the calling function to release the used memory after it has finished processing it.
In this case, you can't tell the logical size of the vector, so you have to set answer[k] to a special value (-1 here), to know where the vector ends.
#include <assert.h>
int* toBase(int n, int b) {
assert(b > 1);
int* answer = malloc(33 * sizeof(int));
q = n
k = 0
while (q != 0) {
answer[k] = q % b;
q /= b;
++k;
}
answer[k] = -1;
return answer;
}
int main() {
int n = 100000;
int b = 2;
int *answer = toBase(n, b);
// use answer
free(answer);
return 0;
}
A more elegant solution (in C++) is to use the STL vector class.
The idea behind this algorithm is that it's creating a list of values ak, ak-1, ak-2, ..., a0. At the very end, it wants to return that list of values and do so in a form that looks like the base-b representation of the number.
For example, if you input 33 into this algorithm and ask for its base-16 representation, the algorithm will set a1 = 2 and a0 = 1. The return value of the algorithm is then the representation 21, which is a1 (2) followed by a0 (1).
The notation they're using is just fancy mathspeak for "return the list of values produced by this algorithm." You can think of it as returning an array of the base-b digits of the number.
Hope this helps!
I want to write a function that takes an array of letters as an argument and a number of those letters to select.
Say you provide an array of 8 letters and want to select 3 letters from that. Then you should get:
8! / ((8 - 3)! * 3!) = 56
Arrays (or words) in return consisting of 3 letters each.
Art of Computer Programming Volume 4: Fascicle 3 has a ton of these that might fit your particular situation better than how I describe.
Gray Codes
An issue that you will come across is of course memory and pretty quickly, you'll have problems by 20 elements in your set -- 20C3 = 1140. And if you want to iterate over the set it's best to use a modified gray code algorithm so you aren't holding all of them in memory. These generate the next combination from the previous and avoid repetitions. There are many of these for different uses. Do we want to maximize the differences between successive combinations? minimize? et cetera.
Some of the original papers describing gray codes:
Some Hamilton Paths and a Minimal Change Algorithm
Adjacent Interchange Combination Generation Algorithm
Here are some other papers covering the topic:
An Efficient Implementation of the Eades, Hickey, Read Adjacent Interchange Combination Generation Algorithm (PDF, with code in Pascal)
Combination Generators
Survey of Combinatorial Gray Codes (PostScript)
An Algorithm for Gray Codes
Chase's Twiddle (algorithm)
Phillip J Chase, `Algorithm 382: Combinations of M out of N Objects' (1970)
The algorithm in C...
Index of Combinations in Lexicographical Order (Buckles Algorithm 515)
You can also reference a combination by its index (in lexicographical order). Realizing that the index should be some amount of change from right to left based on the index we can construct something that should recover a combination.
So, we have a set {1,2,3,4,5,6}... and we want three elements. Let's say {1,2,3} we can say that the difference between the elements is one and in order and minimal. {1,2,4} has one change and is lexicographically number 2. So the number of 'changes' in the last place accounts for one change in the lexicographical ordering. The second place, with one change {1,3,4} has one change but accounts for more change since it's in the second place (proportional to the number of elements in the original set).
The method I've described is a deconstruction, as it seems, from set to the index, we need to do the reverse – which is much trickier. This is how Buckles solves the problem. I wrote some C to compute them, with minor changes – I used the index of the sets rather than a number range to represent the set, so we are always working from 0...n.
Note:
Since combinations are unordered, {1,3,2} = {1,2,3} --we order them to be lexicographical.
This method has an implicit 0 to start the set for the first difference.
Index of Combinations in Lexicographical Order (McCaffrey)
There is another way:, its concept is easier to grasp and program but it's without the optimizations of Buckles. Fortunately, it also does not produce duplicate combinations:
The set that maximizes , where .
For an example: 27 = C(6,4) + C(5,3) + C(2,2) + C(1,1). So, the 27th lexicographical combination of four things is: {1,2,5,6}, those are the indexes of whatever set you want to look at. Example below (OCaml), requires choose function, left to reader:
(* this will find the [x] combination of a [set] list when taking [k] elements *)
let combination_maccaffery set k x =
(* maximize function -- maximize a that is aCb *)
(* return largest c where c < i and choose(c,i) <= z *)
let rec maximize a b x =
if (choose a b ) <= x then a else maximize (a-1) b x
in
let rec iterate n x i = match i with
| 0 -> []
| i ->
let max = maximize n i x in
max :: iterate n (x - (choose max i)) (i-1)
in
if x < 0 then failwith "errors" else
let idxs = iterate (List.length set) x k in
List.map (List.nth set) (List.sort (-) idxs)
A small and simple combinations iterator
The following two algorithms are provided for didactic purposes. They implement an iterator and (a more general) folder overall combinations.
They are as fast as possible, having the complexity O(nCk). The memory consumption is bound by k.
We will start with the iterator, which will call a user provided function for each combination
let iter_combs n k f =
let rec iter v s j =
if j = k then f v
else for i = s to n - 1 do iter (i::v) (i+1) (j+1) done in
iter [] 0 0
A more general version will call the user provided function along with the state variable, starting from the initial state. Since we need to pass the state between different states we won't use the for-loop, but instead, use recursion,
let fold_combs n k f x =
let rec loop i s c x =
if i < n then
loop (i+1) s c ##
let c = i::c and s = s + 1 and i = i + 1 in
if s < k then loop i s c x else f c x
else x in
loop 0 0 [] x
In C#:
public static IEnumerable<IEnumerable<T>> Combinations<T>(this IEnumerable<T> elements, int k)
{
return k == 0 ? new[] { new T[0] } :
elements.SelectMany((e, i) =>
elements.Skip(i + 1).Combinations(k - 1).Select(c => (new[] {e}).Concat(c)));
}
Usage:
var result = Combinations(new[] { 1, 2, 3, 4, 5 }, 3);
Result:
123
124
125
134
135
145
234
235
245
345
Short java solution:
import java.util.Arrays;
public class Combination {
public static void main(String[] args){
String[] arr = {"A","B","C","D","E","F"};
combinations2(arr, 3, 0, new String[3]);
}
static void combinations2(String[] arr, int len, int startPosition, String[] result){
if (len == 0){
System.out.println(Arrays.toString(result));
return;
}
for (int i = startPosition; i <= arr.length-len; i++){
result[result.length - len] = arr[i];
combinations2(arr, len-1, i+1, result);
}
}
}
Result will be
[A, B, C]
[A, B, D]
[A, B, E]
[A, B, F]
[A, C, D]
[A, C, E]
[A, C, F]
[A, D, E]
[A, D, F]
[A, E, F]
[B, C, D]
[B, C, E]
[B, C, F]
[B, D, E]
[B, D, F]
[B, E, F]
[C, D, E]
[C, D, F]
[C, E, F]
[D, E, F]
May I present my recursive Python solution to this problem?
def choose_iter(elements, length):
for i in xrange(len(elements)):
if length == 1:
yield (elements[i],)
else:
for next in choose_iter(elements[i+1:], length-1):
yield (elements[i],) + next
def choose(l, k):
return list(choose_iter(l, k))
Example usage:
>>> len(list(choose_iter("abcdefgh",3)))
56
I like it for its simplicity.
Lets say your array of letters looks like this: "ABCDEFGH". You have three indices (i, j, k) indicating which letters you are going to use for the current word, You start with:
A B C D E F G H
^ ^ ^
i j k
First you vary k, so the next step looks like that:
A B C D E F G H
^ ^ ^
i j k
If you reached the end you go on and vary j and then k again.
A B C D E F G H
^ ^ ^
i j k
A B C D E F G H
^ ^ ^
i j k
Once you j reached G you start also to vary i.
A B C D E F G H
^ ^ ^
i j k
A B C D E F G H
^ ^ ^
i j k
...
Written in code this look something like that
void print_combinations(const char *string)
{
int i, j, k;
int len = strlen(string);
for (i = 0; i < len - 2; i++)
{
for (j = i + 1; j < len - 1; j++)
{
for (k = j + 1; k < len; k++)
printf("%c%c%c\n", string[i], string[j], string[k]);
}
}
}
The following recursive algorithm picks all of the k-element combinations from an ordered set:
choose the first element i of your combination
combine i with each of the combinations of k-1 elements chosen recursively from the set of elements larger than i.
Iterate the above for each i in the set.
It is essential that you pick the rest of the elements as larger than i, to avoid repetition. This way [3,5] will be picked only once, as [3] combined with [5], instead of twice (the condition eliminates [5] + [3]). Without this condition you get variations instead of combinations.
Short example in Python:
def comb(sofar, rest, n):
if n == 0:
print sofar
else:
for i in range(len(rest)):
comb(sofar + rest[i], rest[i+1:], n-1)
>>> comb("", "abcde", 3)
abc
abd
abe
acd
ace
ade
bcd
bce
bde
cde
For explanation, the recursive method is described with the following example:
Example: A B C D E
All combinations of 3 would be:
A with all combinations of 2 from the rest (B C D E)
B with all combinations of 2 from the rest (C D E)
C with all combinations of 2 from the rest (D E)
I found this thread useful and thought I would add a Javascript solution that you can pop into Firebug. Depending on your JS engine, it could take a little time if the starting string is large.
function string_recurse(active, rest) {
if (rest.length == 0) {
console.log(active);
} else {
string_recurse(active + rest.charAt(0), rest.substring(1, rest.length));
string_recurse(active, rest.substring(1, rest.length));
}
}
string_recurse("", "abc");
The output should be as follows:
abc
ab
ac
a
bc
b
c
In C++ the following routine will produce all combinations of length distance(first,k) between the range [first,last):
#include <algorithm>
template <typename Iterator>
bool next_combination(const Iterator first, Iterator k, const Iterator last)
{
/* Credits: Mark Nelson http://marknelson.us */
if ((first == last) || (first == k) || (last == k))
return false;
Iterator i1 = first;
Iterator i2 = last;
++i1;
if (last == i1)
return false;
i1 = last;
--i1;
i1 = k;
--i2;
while (first != i1)
{
if (*--i1 < *i2)
{
Iterator j = k;
while (!(*i1 < *j)) ++j;
std::iter_swap(i1,j);
++i1;
++j;
i2 = k;
std::rotate(i1,j,last);
while (last != j)
{
++j;
++i2;
}
std::rotate(k,i2,last);
return true;
}
}
std::rotate(first,k,last);
return false;
}
It can be used like this:
#include <string>
#include <iostream>
int main()
{
std::string s = "12345";
std::size_t comb_size = 3;
do
{
std::cout << std::string(s.begin(), s.begin() + comb_size) << std::endl;
} while (next_combination(s.begin(), s.begin() + comb_size, s.end()));
return 0;
}
This will print the following:
123
124
125
134
135
145
234
235
245
345
static IEnumerable<string> Combinations(List<string> characters, int length)
{
for (int i = 0; i < characters.Count; i++)
{
// only want 1 character, just return this one
if (length == 1)
yield return characters[i];
// want more than one character, return this one plus all combinations one shorter
// only use characters after the current one for the rest of the combinations
else
foreach (string next in Combinations(characters.GetRange(i + 1, characters.Count - (i + 1)), length - 1))
yield return characters[i] + next;
}
}
Simple recursive algorithm in Haskell
import Data.List
combinations 0 lst = [[]]
combinations n lst = do
(x:xs) <- tails lst
rest <- combinations (n-1) xs
return $ x : rest
We first define the special case, i.e. selecting zero elements. It produces a single result, which is an empty list (i.e. a list that contains an empty list).
For n > 0, x goes through every element of the list and xs is every element after x.
rest picks n - 1 elements from xs using a recursive call to combinations. The final result of the function is a list where each element is x : rest (i.e. a list which has x as head and rest as tail) for every different value of x and rest.
> combinations 3 "abcde"
["abc","abd","abe","acd","ace","ade","bcd","bce","bde","cde"]
And of course, since Haskell is lazy, the list is gradually generated as needed, so you can partially evaluate exponentially large combinations.
> let c = combinations 8 "abcdefghijklmnopqrstuvwxyz"
> take 10 c
["abcdefgh","abcdefgi","abcdefgj","abcdefgk","abcdefgl","abcdefgm","abcdefgn",
"abcdefgo","abcdefgp","abcdefgq"]
And here comes granddaddy COBOL, the much maligned language.
Let's assume an array of 34 elements of 8 bytes each (purely arbitrary selection.) The idea is to enumerate all possible 4-element combinations and load them into an array.
We use 4 indices, one each for each position in the group of 4
The array is processed like this:
idx1 = 1
idx2 = 2
idx3 = 3
idx4 = 4
We vary idx4 from 4 to the end. For each idx4 we get a unique combination
of groups of four. When idx4 comes to the end of the array, we increment idx3 by 1 and set idx4 to idx3+1. Then we run idx4 to the end again. We proceed in this manner, augmenting idx3,idx2, and idx1 respectively until the position of idx1 is less than 4 from the end of the array. That finishes the algorithm.
1 --- pos.1
2 --- pos 2
3 --- pos 3
4 --- pos 4
5
6
7
etc.
First iterations:
1234
1235
1236
1237
1245
1246
1247
1256
1257
1267
etc.
A COBOL example:
01 DATA_ARAY.
05 FILLER PIC X(8) VALUE "VALUE_01".
05 FILLER PIC X(8) VALUE "VALUE_02".
etc.
01 ARAY_DATA OCCURS 34.
05 ARAY_ITEM PIC X(8).
01 OUTPUT_ARAY OCCURS 50000 PIC X(32).
01 MAX_NUM PIC 99 COMP VALUE 34.
01 INDEXXES COMP.
05 IDX1 PIC 99.
05 IDX2 PIC 99.
05 IDX3 PIC 99.
05 IDX4 PIC 99.
05 OUT_IDX PIC 9(9).
01 WHERE_TO_STOP_SEARCH PIC 99 COMP.
* Stop the search when IDX1 is on the third last array element:
COMPUTE WHERE_TO_STOP_SEARCH = MAX_VALUE - 3
MOVE 1 TO IDX1
PERFORM UNTIL IDX1 > WHERE_TO_STOP_SEARCH
COMPUTE IDX2 = IDX1 + 1
PERFORM UNTIL IDX2 > MAX_NUM
COMPUTE IDX3 = IDX2 + 1
PERFORM UNTIL IDX3 > MAX_NUM
COMPUTE IDX4 = IDX3 + 1
PERFORM UNTIL IDX4 > MAX_NUM
ADD 1 TO OUT_IDX
STRING ARAY_ITEM(IDX1)
ARAY_ITEM(IDX2)
ARAY_ITEM(IDX3)
ARAY_ITEM(IDX4)
INTO OUTPUT_ARAY(OUT_IDX)
ADD 1 TO IDX4
END-PERFORM
ADD 1 TO IDX3
END-PERFORM
ADD 1 TO IDX2
END_PERFORM
ADD 1 TO IDX1
END-PERFORM.
Another C# version with lazy generation of the combination indices. This version maintains a single array of indices to define a mapping between the list of all values and the values for the current combination, i.e. constantly uses O(k) additional space during the entire runtime. The code generates individual combinations, including the first one, in O(k) time.
public static IEnumerable<T[]> Combinations<T>(this T[] values, int k)
{
if (k < 0 || values.Length < k)
yield break; // invalid parameters, no combinations possible
// generate the initial combination indices
var combIndices = new int[k];
for (var i = 0; i < k; i++)
{
combIndices[i] = i;
}
while (true)
{
// return next combination
var combination = new T[k];
for (var i = 0; i < k; i++)
{
combination[i] = values[combIndices[i]];
}
yield return combination;
// find first index to update
var indexToUpdate = k - 1;
while (indexToUpdate >= 0 && combIndices[indexToUpdate] >= values.Length - k + indexToUpdate)
{
indexToUpdate--;
}
if (indexToUpdate < 0)
yield break; // done
// update combination indices
for (var combIndex = combIndices[indexToUpdate] + 1; indexToUpdate < k; indexToUpdate++, combIndex++)
{
combIndices[indexToUpdate] = combIndex;
}
}
}
Test code:
foreach (var combination in new[] {'a', 'b', 'c', 'd', 'e'}.Combinations(3))
{
System.Console.WriteLine(String.Join(" ", combination));
}
Output:
a b c
a b d
a b e
a c d
a c e
a d e
b c d
b c e
b d e
c d e
Here is an elegant, generic implementation in Scala, as described on 99 Scala Problems.
object P26 {
def flatMapSublists[A,B](ls: List[A])(f: (List[A]) => List[B]): List[B] =
ls match {
case Nil => Nil
case sublist#(_ :: tail) => f(sublist) ::: flatMapSublists(tail)(f)
}
def combinations[A](n: Int, ls: List[A]): List[List[A]] =
if (n == 0) List(Nil)
else flatMapSublists(ls) { sl =>
combinations(n - 1, sl.tail) map {sl.head :: _}
}
}
If you can use SQL syntax - say, if you're using LINQ to access fields of an structure or array, or directly accessing a database that has a table called "Alphabet" with just one char field "Letter", you can adapt following code:
SELECT A.Letter, B.Letter, C.Letter
FROM Alphabet AS A, Alphabet AS B, Alphabet AS C
WHERE A.Letter<>B.Letter AND A.Letter<>C.Letter AND B.Letter<>C.Letter
AND A.Letter<B.Letter AND B.Letter<C.Letter
This will return all combinations of 3 letters, notwithstanding how many letters you have in table "Alphabet" (it can be 3, 8, 10, 27, etc.).
If what you want is all permutations, rather than combinations (i.e. you want "ACB" and "ABC" to count as different, rather than appear just once) just delete the last line (the AND one) and it's done.
Post-Edit: After re-reading the question, I realise what's needed is the general algorithm, not just a specific one for the case of selecting 3 items. Adam Hughes' answer is the complete one, unfortunately I cannot vote it up (yet). This answer's simple but works only for when you want exactly 3 items.
I had a permutation algorithm I used for project euler, in python:
def missing(miss,src):
"Returns the list of items in src not present in miss"
return [i for i in src if i not in miss]
def permutation_gen(n,l):
"Generates all the permutations of n items of the l list"
for i in l:
if n<=1: yield [i]
r = [i]
for j in permutation_gen(n-1,missing([i],l)): yield r+j
If
n<len(l)
you should have all combination you need without repetition, do you need it?
It is a generator, so you use it in something like this:
for comb in permutation_gen(3,list("ABCDEFGH")):
print comb
https://gist.github.com/3118596
There is an implementation for JavaScript. It has functions to get k-combinations and all combinations of an array of any objects. Examples:
k_combinations([1,2,3], 2)
-> [[1,2], [1,3], [2,3]]
combinations([1,2,3])
-> [[1],[2],[3],[1,2],[1,3],[2,3],[1,2,3]]
Lets say your array of letters looks like this: "ABCDEFGH". You have three indices (i, j, k) indicating which letters you are going to use for the current word, You start with:
A B C D E F G H
^ ^ ^
i j k
First you vary k, so the next step looks like that:
A B C D E F G H
^ ^ ^
i j k
If you reached the end you go on and vary j and then k again.
A B C D E F G H
^ ^ ^
i j k
A B C D E F G H
^ ^ ^
i j k
Once you j reached G you start also to vary i.
A B C D E F G H
^ ^ ^
i j k
A B C D E F G H
^ ^ ^
i j k
...
function initializePointers($cnt) {
$pointers = [];
for($i=0; $i<$cnt; $i++) {
$pointers[] = $i;
}
return $pointers;
}
function incrementPointers(&$pointers, &$arrLength) {
for($i=0; $i<count($pointers); $i++) {
$currentPointerIndex = count($pointers) - $i - 1;
$currentPointer = $pointers[$currentPointerIndex];
if($currentPointer < $arrLength - $i - 1) {
++$pointers[$currentPointerIndex];
for($j=1; ($currentPointerIndex+$j)<count($pointers); $j++) {
$pointers[$currentPointerIndex+$j] = $pointers[$currentPointerIndex]+$j;
}
return true;
}
}
return false;
}
function getDataByPointers(&$arr, &$pointers) {
$data = [];
for($i=0; $i<count($pointers); $i++) {
$data[] = $arr[$pointers[$i]];
}
return $data;
}
function getCombinations($arr, $cnt)
{
$len = count($arr);
$result = [];
$pointers = initializePointers($cnt);
do {
$result[] = getDataByPointers($arr, $pointers);
} while(incrementPointers($pointers, count($arr)));
return $result;
}
$result = getCombinations([0, 1, 2, 3, 4, 5], 3);
print_r($result);
Based on https://stackoverflow.com/a/127898/2628125, but more abstract, for any size of pointers.
Here you have a lazy evaluated version of that algorithm coded in C#:
static bool nextCombination(int[] num, int n, int k)
{
bool finished, changed;
changed = finished = false;
if (k > 0)
{
for (int i = k - 1; !finished && !changed; i--)
{
if (num[i] < (n - 1) - (k - 1) + i)
{
num[i]++;
if (i < k - 1)
{
for (int j = i + 1; j < k; j++)
{
num[j] = num[j - 1] + 1;
}
}
changed = true;
}
finished = (i == 0);
}
}
return changed;
}
static IEnumerable Combinations<T>(IEnumerable<T> elements, int k)
{
T[] elem = elements.ToArray();
int size = elem.Length;
if (k <= size)
{
int[] numbers = new int[k];
for (int i = 0; i < k; i++)
{
numbers[i] = i;
}
do
{
yield return numbers.Select(n => elem[n]);
}
while (nextCombination(numbers, size, k));
}
}
And test part:
static void Main(string[] args)
{
int k = 3;
var t = new[] { "dog", "cat", "mouse", "zebra"};
foreach (IEnumerable<string> i in Combinations(t, k))
{
Console.WriteLine(string.Join(",", i));
}
}
Hope this help you!
Another version, that forces all the first k to appear firstly, then all the first k+1 combinations, then all the first k+2 etc.. It means that if you have sorted array, the most important on the top, it would take them and expand gradually to the next ones - only when it is must do so.
private static bool NextCombinationFirstsAlwaysFirst(int[] num, int n, int k)
{
if (k > 1 && NextCombinationFirstsAlwaysFirst(num, num[k - 1], k - 1))
return true;
if (num[k - 1] + 1 == n)
return false;
++num[k - 1];
for (int i = 0; i < k - 1; ++i)
num[i] = i;
return true;
}
For instance, if you run the first method ("nextCombination") on k=3, n=5 you'll get:
0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4
But if you'll run
int[] nums = new int[k];
for (int i = 0; i < k; ++i)
nums[i] = i;
do
{
Console.WriteLine(string.Join(" ", nums));
}
while (NextCombinationFirstsAlwaysFirst(nums, n, k));
You'll get this (I added empty lines for clarity):
0 1 2
0 1 3
0 2 3
1 2 3
0 1 4
0 2 4
1 2 4
0 3 4
1 3 4
2 3 4
It's adding "4" only when must to, and also after "4" was added it adds "3" again only when it must to (after doing 01, 02, 12).
Array.prototype.combs = function(num) {
var str = this,
length = str.length,
of = Math.pow(2, length) - 1,
out, combinations = [];
while(of) {
out = [];
for(var i = 0, y; i < length; i++) {
y = (1 << i);
if(y & of && (y !== of))
out.push(str[i]);
}
if (out.length >= num) {
combinations.push(out);
}
of--;
}
return combinations;
}
Clojure version:
(defn comb [k l]
(if (= 1 k) (map vector l)
(apply concat
(map-indexed
#(map (fn [x] (conj x %2))
(comb (dec k) (drop (inc %1) l)))
l))))
Algorithm:
Count from 1 to 2^n.
Convert each digit to its binary representation.
Translate each 'on' bit to elements of your set, based on position.
In C#:
void Main()
{
var set = new [] {"A", "B", "C", "D" }; //, "E", "F", "G", "H", "I", "J" };
var kElement = 2;
for(var i = 1; i < Math.Pow(2, set.Length); i++) {
var result = Convert.ToString(i, 2).PadLeft(set.Length, '0');
var cnt = Regex.Matches(Regex.Escape(result), "1").Count;
if (cnt == kElement) {
for(int j = 0; j < set.Length; j++)
if ( Char.GetNumericValue(result[j]) == 1)
Console.Write(set[j]);
Console.WriteLine();
}
}
}
Why does it work?
There is a bijection between the subsets of an n-element set and n-bit sequences.
That means we can figure out how many subsets there are by counting sequences.
e.g., the four element set below can be represented by {0,1} X {0, 1} X {0, 1} X {0, 1} (or 2^4) different sequences.
So - all we have to do is count from 1 to 2^n to find all the combinations. (We ignore the empty set.) Next, translate the digits to their binary representation. Then substitute elements of your set for 'on' bits.
If you want only k element results, only print when k bits are 'on'.
(If you want all subsets instead of k length subsets, remove the cnt/kElement part.)
(For proof, see MIT free courseware Mathematics for Computer Science, Lehman et al, section 11.2.2. https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/readings/ )
short python code, yielding index positions
def yield_combos(n,k):
# n is set size, k is combo size
i = 0
a = [0]*k
while i > -1:
for j in range(i+1, k):
a[j] = a[j-1]+1
i=j
yield a
while a[i] == i + n - k:
i -= 1
a[i] += 1
All said and and done here comes the O'caml code for that.
Algorithm is evident from the code..
let combi n lst =
let rec comb l c =
if( List.length c = n) then [c] else
match l with
[] -> []
| (h::t) -> (combi t (h::c))#(combi t c)
in
combi lst []
;;
Here is a method which gives you all combinations of specified size from a random length string. Similar to quinmars' solution, but works for varied input and k.
The code can be changed to wrap around, ie 'dab' from input 'abcd' w k=3.
public void run(String data, int howMany){
choose(data, howMany, new StringBuffer(), 0);
}
//n choose k
private void choose(String data, int k, StringBuffer result, int startIndex){
if (result.length()==k){
System.out.println(result.toString());
return;
}
for (int i=startIndex; i<data.length(); i++){
result.append(data.charAt(i));
choose(data,k,result, i+1);
result.setLength(result.length()-1);
}
}
Output for "abcde":
abc abd abe acd ace ade bcd bce bde cde
Short javascript version (ES 5)
let combine = (list, n) =>
n == 0 ?
[[]] :
list.flatMap((e, i) =>
combine(
list.slice(i + 1),
n - 1
).map(c => [e].concat(c))
);
let res = combine([1,2,3,4], 3);
res.forEach(e => console.log(e.join()));
Another python recusive solution.
def combination_indicies(n, k, j = 0, stack = []):
if len(stack) == k:
yield list(stack)
return
for i in range(j, n):
stack.append(i)
for x in combination_indicies(n, k, i + 1, stack):
yield x
stack.pop()
list(combination_indicies(5, 3))
Output:
[[0, 1, 2],
[0, 1, 3],
[0, 1, 4],
[0, 2, 3],
[0, 2, 4],
[0, 3, 4],
[1, 2, 3],
[1, 2, 4],
[1, 3, 4],
[2, 3, 4]]
I created a solution in SQL Server 2005 for this, and posted it on my website: http://www.jessemclain.com/downloads/code/sql/fn_GetMChooseNCombos.sql.htm
Here is an example to show usage:
SELECT * FROM dbo.fn_GetMChooseNCombos('ABCD', 2, '')
results:
Word
----
AB
AC
AD
BC
BD
CD
(6 row(s) affected)
Here is my proposition in C++
I tried to impose as little restriction on the iterator type as i could so this solution assumes just forward iterator, and it can be a const_iterator. This should work with any standard container. In cases where arguments don't make sense it throws std::invalid_argumnent
#include <vector>
#include <stdexcept>
template <typename Fci> // Fci - forward const iterator
std::vector<std::vector<Fci> >
enumerate_combinations(Fci begin, Fci end, unsigned int combination_size)
{
if(begin == end && combination_size > 0u)
throw std::invalid_argument("empty set and positive combination size!");
std::vector<std::vector<Fci> > result; // empty set of combinations
if(combination_size == 0u) return result; // there is exactly one combination of
// size 0 - emty set
std::vector<Fci> current_combination;
current_combination.reserve(combination_size + 1u); // I reserve one aditional slot
// in my vector to store
// the end sentinel there.
// The code is cleaner thanks to that
for(unsigned int i = 0u; i < combination_size && begin != end; ++i, ++begin)
{
current_combination.push_back(begin); // Construction of the first combination
}
// Since I assume the itarators support only incrementing, I have to iterate over
// the set to get its size, which is expensive. Here I had to itrate anyway to
// produce the first cobination, so I use the loop to also check the size.
if(current_combination.size() < combination_size)
throw std::invalid_argument("combination size > set size!");
result.push_back(current_combination); // Store the first combination in the results set
current_combination.push_back(end); // Here I add mentioned earlier sentinel to
// simplyfy rest of the code. If I did it
// earlier, previous statement would get ugly.
while(true)
{
unsigned int i = combination_size;
Fci tmp; // Thanks to the sentinel I can find first
do // iterator to change, simply by scaning
{ // from right to left and looking for the
tmp = current_combination[--i]; // first "bubble". The fact, that it's
++tmp; // a forward iterator makes it ugly but I
} // can't help it.
while(i > 0u && tmp == current_combination[i + 1u]);
// Here is probably my most obfuscated expression.
// Loop above looks for a "bubble". If there is no "bubble", that means, that
// current_combination is the last combination, Expression in the if statement
// below evaluates to true and the function exits returning result.
// If the "bubble" is found however, the ststement below has a sideeffect of
// incrementing the first iterator to the left of the "bubble".
if(++current_combination[i] == current_combination[i + 1u])
return result;
// Rest of the code sets posiotons of the rest of the iterstors
// (if there are any), that are to the right of the incremented one,
// to form next combination
while(++i < combination_size)
{
current_combination[i] = current_combination[i - 1u];
++current_combination[i];
}
// Below is the ugly side of using the sentinel. Well it had to haave some
// disadvantage. Try without it.
result.push_back(std::vector<Fci>(current_combination.begin(),
current_combination.end() - 1));
}
}
Here is a code I recently wrote in Java, which calculates and returns all the combination of "num" elements from "outOf" elements.
// author: Sourabh Bhat (heySourabh#gmail.com)
public class Testing
{
public static void main(String[] args)
{
// Test case num = 5, outOf = 8.
int num = 5;
int outOf = 8;
int[][] combinations = getCombinations(num, outOf);
for (int i = 0; i < combinations.length; i++)
{
for (int j = 0; j < combinations[i].length; j++)
{
System.out.print(combinations[i][j] + " ");
}
System.out.println();
}
}
private static int[][] getCombinations(int num, int outOf)
{
int possibilities = get_nCr(outOf, num);
int[][] combinations = new int[possibilities][num];
int arrayPointer = 0;
int[] counter = new int[num];
for (int i = 0; i < num; i++)
{
counter[i] = i;
}
breakLoop: while (true)
{
// Initializing part
for (int i = 1; i < num; i++)
{
if (counter[i] >= outOf - (num - 1 - i))
counter[i] = counter[i - 1] + 1;
}
// Testing part
for (int i = 0; i < num; i++)
{
if (counter[i] < outOf)
{
continue;
} else
{
break breakLoop;
}
}
// Innermost part
combinations[arrayPointer] = counter.clone();
arrayPointer++;
// Incrementing part
counter[num - 1]++;
for (int i = num - 1; i >= 1; i--)
{
if (counter[i] >= outOf - (num - 1 - i))
counter[i - 1]++;
}
}
return combinations;
}
private static int get_nCr(int n, int r)
{
if(r > n)
{
throw new ArithmeticException("r is greater then n");
}
long numerator = 1;
long denominator = 1;
for (int i = n; i >= r + 1; i--)
{
numerator *= i;
}
for (int i = 2; i <= n - r; i++)
{
denominator *= i;
}
return (int) (numerator / denominator);
}
}