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I have an array of numbers and I want to figure out the maximum length of a continuous subarray of 2 unique numbers repeating.
For example, [2, 3, 4, 3, 2, 2, 4] would return 3 since [3, 2, 2] is of length 3.
[2, 4, 2, 5, 1, 5, 4, 2] would return 3.
[7, 8, 7, 8, 7] would return 5.
Edit: I have considered an O(n^2) solution where I start at each value in the array and iterate until I see a third unique value.
for item in array:
iterate until third unique element
if length of this iteration is greater than existing max, update the max length
return maxlength
I do not, however, think this is an efficient solution.
It can be done O(n). The code is in python3. o and t are one and two respectively. m is the max and c is the current count variable.
a = [7, 8, 7, 8, 7]
m = -1
o = a[0]
t = a[1]
# in the beginning one and two are the first 2 numbers
c = 0
index = 0
for i in a:
if i == o or i == t:
# if current element is either one or two current count is increased
c += 1
else:
# if current element is neither one nor two then they are updated accordingly and max is updated
o = a[index - 1]
t = a[index]
m = max(m, c)
c = 2
index += 1
m = max(m, c)
print(m)
We can use two pointer technique to solve this problem in O(n) run time complexity. These two pointer for example startPtr and endPtr will represent the range in the array. We will maintain this range [startPtr, endPtr] in such way that it contains no more than 2 unique number. We can do this by keeping track of position of the 2 unique number. My implement in C++ is given below:
int main()
{
int array[] = {1,2,3,3,2,3,2,3,2,2,2,1,3,4};
int startPtr = 0;
int endPtr = 0;
// denotes the size of the array
int size= sizeof(array)/sizeof(array[0]);
// contain last position of unique number 1 in the range [startPtr, endPtr]
int uniqueNumPos1 = -1; // -1 value represents it is not set yet
// contain last position of unique number 2 in the range [startPtr, endPtr]
int uniqueNumPos2 = -1; // -1 value represents it is not set yet
// contains length of maximum continuous subarray with 2 unique numbers
int ans = 0;
while(endPtr < size) {
if(uniqueNumPos1 == -1 || array[endPtr] == array[uniqueNumPos1]) {
uniqueNumPos1 = endPtr;
}
else {
if(uniqueNumPos2 == -1 || array[endPtr] == array[uniqueNumPos2]) {
uniqueNumPos2 = endPtr;
}
else {
// for this new third unique number update startPtr with min(uniqueNumPos1, uniqueNumPos2) + 1
// to ensure [startPtr, endPtr] does not contain more that two unique
startPtr = min(uniqueNumPos1, uniqueNumPos2) + 1;
// update uniqueNumPos1 and uniqueNumPos2
uniqueNumPos1 = endPtr -1;
uniqueNumPos2 = endPtr;
}
}
// this conditon is to ensure the range contain exactly two unique number
// if you are looking for the range containing less than or equal to two unique number, then you can omit this condition
if (uniqueNumPos1 != -1 && uniqueNumPos2 !=-1) {
ans = max( ans, endPtr - startPtr + 1);
}
endPtr++;
}
printf("%d\n", ans);
}
Thanks #MBo for pointing out the mistakes.
import java.util.Arrays;
import static java.lang.System.out;
class TestCase{
int[] test;
int answer;
TestCase(int[] test,int answer){
this.test = test;
this.answer = answer;
}
}
public class Solution {
public static void main(String[] args) {
TestCase[] tests = {
new TestCase(new int[]{2, 3, 4, 3, 2, 2, 4},3),
new TestCase(new int[]{2, 3, 3, 3, 3, 4, 3, 3, 2, 2, 4},7),
new TestCase(new int[]{1,2,3,3,4,2,3,2,3,2,2,2,1,3,4},7),
new TestCase(new int[]{2, 7, 8, 7, 8, 7},5),
new TestCase(new int[]{-1,2,2,2,2,2,2,2,2,2,2,-1,-1,4},13),
new TestCase(new int[]{1,2,3,4,5,6,7,7},3),
new TestCase(new int[]{0,0,0,0,0},0),
new TestCase(new int[]{0,0,0,2,2,2,1,1,1,1},7),
new TestCase(new int[]{},0)
};
for(int i=0;i<tests.length;++i){
int ans = maxContiguousArrayWith2UniqueElements(tests[i].test);
out.println(Arrays.toString(tests[i].test));
out.println("Expected: " + tests[i].answer);
out.println("Returned: " + ans);
out.println("Result: " + (tests[i].answer == ans ? "ok" : "not ok"));
out.println();
}
}
private static int maxContiguousArrayWith2UniqueElements(int[] A){
if(A == null || A.length <= 1) return 0;
int max_subarray = 0;
int first_number = A[0],second_number = A[0];
int start_index = 0,same_element_run_length = 1;
for(int i=1;i<A.length;++i){
if(A[i] != A[i-1]){
if(first_number == second_number){
second_number = A[i];
}else{
if(A[i] != first_number && A[i] != second_number){
max_subarray = Math.max(max_subarray,i - start_index);
start_index = i - same_element_run_length;
first_number = A[i-1];
second_number = A[i];
}
}
same_element_run_length = 1;
}else{
same_element_run_length++;
}
}
return first_number == second_number ? max_subarray : Math.max(max_subarray,A.length - start_index);
}
}
OUTPUT:
[2, 3, 4, 3, 2, 2, 4]
Expected: 3
Returned: 3
Result: ok
[2, 3, 3, 3, 3, 4, 3, 3, 2, 2, 4]
Expected: 7
Returned: 7
Result: ok
[1, 2, 3, 3, 4, 2, 3, 2, 3, 2, 2, 2, 1, 3, 4]
Expected: 7
Returned: 7
Result: ok
[2, 7, 8, 7, 8, 7]
Expected: 5
Returned: 5
Result: ok
[-1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, -1, -1, 4]
Expected: 13
Returned: 13
Result: ok
[1, 2, 3, 4, 5, 6, 7, 7]
Expected: 3
Returned: 3
Result: ok
[0, 0, 0, 0, 0]
Expected: 0
Returned: 0
Result: ok
[0, 0, 0, 2, 2, 2, 1, 1, 1, 1]
Expected: 7
Returned: 7
Result: ok
[]
Expected: 0
Returned: 0
Result: ok
Algorithm:
So, we maintain 2 variables first_number and second_number which will hold those 2 unique numbers.
As you know, there could be many possible subarrays we have to consider to get the max subarray length which has 2 unique elements. Hence, we need a pointer variable which will point to start of a subarray. In this, that pointer is start_index.
Any subarray breaks when we find a third number which is not equal to first_number or second_number. So, now, we calculate the previous subarray length(which had those 2 unique elements) by doing i - start_index.
Tricky part of this question is how to get the start_index of the next subarray.
If you closely observe, second_number of previous subarray becomes first_number of current subarray and third number we encountered just now becomes second_number of this current subarray.
So, one way to calculate when this first_number started is to run a while loop backwards to get that start_index. But that would make the algorithm O(n^2) if there are many subarrays to consider(which it will be).
Hence, we maintain a variable called same_element_run_length which just keeps track of the length or frequency of how many times the number got repeated and gets updated whenever it breaks. So, start_index for the next subarray after we encounter the third number becomes start_index = i - same_element_run_length.
Rest of the computation done is self-explanatory.
Time Complexity: O(n), Space Complexity : O(1).
For example, if A={0,1,2,3,4}, r=3 and B={1,4}, the result would be:
[0, 1, 2]
[0, 1, 3]
[0, 1, 4]
[0, 2, 4]
[0, 3, 4]
[1, 2, 3]
[1, 2, 4]
[1, 3, 4]
[2, 3, 4]
That's all the r-long combinations of A, excluding [0, 2, 3], because that one doesn't contain either 1 or 4.
The solution that I currently have is the following, using the fastest algorithm for getting normal combinations I know of, and just doing a simple check to see if combinations generated also contain an element of B (java):
int[] A = new int[]{0,1,2,3,4};
int[] B = new int[]{1,4};
int n = A.length;
int r = 3;
int[] picks = new int[r]; //Holds indexes of elements in A
for (int i = 0; i < picks.length; i++)
picks[i] = i;
int lastindex = picks.length - 1;
outer:
while (true) {
int at = lastindex;
while (true) {
picks[at] += 1;
if (picks[at] < n) {
int displacement = picks[at] - at; // at + displacement = picks[at], at + displacement + 1 = picks[at] + 1 ,...
// Make all picks elements after at y = picks[at] + x, so picks={0, 2, 4, 6, 18, 30} & at=3 --> picks={0, 2, 4, 5, 6, 7}
// (Note that this example will never take place in reality, because the 18 or the 30 would be increased instead, depending on what n is)
// Do the last one first, because that one is going to be the biggest,
picks[lastindex] = lastindex + displacement;
if (picks[lastindex] < n) { // and check if it doesn't overflow
for (int i = at + 1; i < lastindex; i++)
picks[i] = i + displacement;
int[] combination = new int[r];
for (int i = 0; i < r; i++)
combination[i] = A[picks[i]];
System.out.print(Arrays.toString(combination));
//^With this, all r-long combinations of A get printed
//Straightforward, bruteforce-ish way of checking if int[] combination
//contains any element from B
presence:
for (int p : combination) {
for (int b : B) {
if (p==b) {
System.out.print(" <-- Also contains an element from B");
break presence;
}
}
}
System.out.println();
break;
}
}
at--;
if (at < 0) {
//Moving this check to the start of the while loop will make this natively support pick 0 cases (5C0 for example),
//but reduce performance by I believe quite a bit. Probably better to special-case those (I haven't
// done that in this test tho)
break outer;
}
}
}
output:
[0, 1, 3] <-- Also contains an element from B
[0, 1, 4] <-- Also contains an element from B
[0, 2, 3]
[0, 2, 4] <-- Also contains an element from B
[0, 3, 4] <-- Also contains an element from B
[1, 2, 3] <-- Also contains an element from B
[1, 2, 4] <-- Also contains an element from B
[1, 3, 4] <-- Also contains an element from B
[2, 3, 4] <-- Also contains an element from B
As written in the comments, I believe this method to be very rudimentary. Can anyone think of a faster way to do this?
Assuming you have a int[][] FindCombinations(int[] set, int length) function that returns a list of all the length-long combinations of set, do the following (pseudo-code):
for i=1 to B.length
{
int bi = B[i];
A = A - bi; // remove bi from A
foreach C in FindCombinations(A, r-1)
{
output C+bi // output the union of C and {bi}
}
}
This way all combinations contain at least one element from B (and may also contain elements of B that have not yet been used) without much extra work. All other combinations are eliminated at no cost (the don't have to be found at all) and also the test that a combination contains an element from B for each combination is also eliminated.
Whether this algorithm is faster, greatly depends on how efficently you can add/remove elements from a set and the percentage of included vs excluded combinations (i.e. if you only end up excluding 1% of the total combinations it is probably not worth it)
Note that when getting the combinations to union with {b[i]} these may also contain an element B[j] where j>i. When you get to the point that you get the combinations to union with B[j] none of them will contain B[i], so all combinations are unique.
This is somewhat of a combinatorial problem; I'm trying to figure out an efficient way to pair up all items in a data set.
For example, I have an array of length 6: [1,2,3,4,5,6], and I want to make all possible pairings of the contents in the array as so:
[1,2],[3,4],[5,6]
[1,2],[3,5],[4,6]
[1,2],[3,6],[4,5]
[1,3],[2,4],[5,6]
[1,3],[2,5],[4,6]
...
and so on. In this example, there would be 15 combinations total. In general, the number of possibilities in this solution should be (N-1)!! where N is the length of the array or the number of items to be paired up. N will always be even in this case. Ideally, an algorithm will generate all possibilities serially, without having to store very large arrays.
For example, one way would work somewhat like a 'round robin' scheduling algorithm where you split the array into N/2:
[1,2,3]
[4,5,6]
and rotate the [5,3,6] clockwise to generate 3 unique pairings (counting vertically) with [1,2,4] fixed as so:
[1,2,3]
[4,5,6]
[1,2,5]
[4,6,3]
[1,2,6]
[4,3,5]
and then rotate [4,2,3,6,5] clockwise to generate 5 unique pairings with [1] fixed, going from the smallest innermost case (N=4) outwards until the end, but recursively. As it is I'm not sure how to best express this as code or if there is a more efficient way of doing this, as N can be very large.
You can generate the pairs using the standard recursive algorithm for generating permutations of a list, but with a twist that each time you recurse you advance 2 elements along the list, and the first remaining element in the list is always the first element of the pair you output at each recursion, with the second of the pair being each of the other remaining elements.
Always choosing the first remaining element as the first item in the pair avoids generating the same pairings with the pairs in different order.
As with the standard algorithm, you can generate the permutations in place without making copies of the array, by swapping elements in place, recursing and then swapping back.
Here is some C code to demonstrate the algorithm (I realise this question is tagged Fortran but just think of it as pseudo code). This code passes the list and count when it recurses, but you could implement it with items and itemcount as global variables:
// start is the current position in the list, advancing by 2 each time
// pass 0 as start when calling at the top level
void generatePairings(int* items, int itemcount, int start)
{
if(itemcount & 1)
return; // must be an even number of items
// is this a complete pairing?
if(start == itemcount)
{
// output pairings:
int i;
for(i = 0; i<itemcount; i+=2)
{
printf("[%d, %d] ", items[i], items[i+1]);
}
printf("\n");
return;
}
// for the next pair, choose the first element in the list for the
// first item in the pair (meaning we don't have to do anything
// but leave it in place), and each of the remaining elements for
// the second item:
int j;
for(j = start+1; j<itemcount; j++)
{
// swap start+1 and j:
int temp = items[start+1];
items[start+1] = items[j];
items[j] = temp;
// recurse:
generatePairings(items, itemcount, start+2);
// swap them back:
temp = items[start+1];
items[start+1] = items[j];
items[j] = temp;
}
}
int main(void) {
int items[6] = {1, 2, 3, 4, 5, 6};
generatePairings(items, 6, 0);
return 0;
}
Output:
[1, 2] [3, 4] [5, 6]
[1, 2] [3, 5] [4, 6]
[1, 2] [3, 6] [5, 4]
[1, 3] [2, 4] [5, 6]
[1, 3] [2, 5] [4, 6]
[1, 3] [2, 6] [5, 4]
[1, 4] [3, 2] [5, 6]
[1, 4] [3, 5] [2, 6]
[1, 4] [3, 6] [5, 2]
[1, 5] [3, 4] [2, 6]
[1, 5] [3, 2] [4, 6]
[1, 5] [3, 6] [2, 4]
[1, 6] [3, 4] [5, 2]
[1, 6] [3, 5] [4, 2]
[1, 6] [3, 2] [5, 4]
If you are doing this on a list of large objects, it's more efficient to permute a list of indices and then use those to index into your array of objects, rather than doing expensive swap operations on large amounts of data.
Wow. Now there's a blast from the past. I wrote about this back in 1993 and provided Pascal source code for it. Surprisingly, the article in which it appeared is available online at http://www.drdobbs.com/database/algorithm-alley/184409099#02e5_000d.
Basically, I adapted a selection sort algorithm:
for x = 0 to n-2
for y = x+1 to n-1
write x, y
The problem with that approach is that it generates {1,2},{1,3},{1,4},{2,3},{2,4}...
It turns out that you can modify that output by maintaining a swap array that you manipulate after every iteration of the outer loop. Here's the Pascal source code that I supplied with the article.
(* ----------------------------------------------------------- *(
** pairings.pas -- Select sports-event team pairings **
** ------------------------------------------------------------**
** This program generates team pairings for sports events. **
** Each team is guaranteed to play each other team exactly **
** once. No team will play more than one game per day. **
** An asterisk ('*') means a day off for that team. **
** For example, 5 teams produces this output: **
** Day 1 - 12 34 5* **
** Day 2 - 13 25 4* **
** Day 3 - 14 2* 35 **
** Day 4 - 15 3* 24 **
** Day 5 - 1* 45 23 **
** ------------------------------------------------------------**
** Copyright (c) 1993 by Jim Mischel. All rights reserved. **
)* ----------------------------------------------------------- *)
program pairings;
const
TEAMCOUNT = 5;
var
TeamNames: Array [1 .. TEAMCOUNT + 1] of Char;
SwapArray: Array [1 .. TEAMCOUNT + 1] of Integer;
x, Temp, Day: Integer;
TempChar: Char;
const
NTeams: Integer = TEAMCOUNT;
begin
{ Set up team names. Normally read from a file. }
for x := 1 to NTeams do
TeamNames[x] := Chr(x + Ord('0'));
if Odd(NTeams) then
begin
NTeams := NTeams + 1;
TeamNames[NTeams] := '*'
end;
{ Set up the array that controls swapping. }
for x := 1 to NTeams do
SwapArray[x] := x;
for Day := 1 to NTeams - 1 do
begin
Write('Day ', Day, ' -');
{ Write the team pairings for this day }
x := 1;
while x < NTeams do
begin
Write(' ', TeamNames[x], TeamNames[x + 1]);
x := x + 2;
end;
WriteLn;
{ Perform swaps to prepare array for next day's pairings. }
if Odd(Day)
then x := 2
else x := 3;
while x < NTeams do
begin
TempChar := TeamNames[SwapArray[x]];
TeamNames[SwapArray[x]] := TeamNames[SwapArray[x + 1]];
TeamNames[SwapArray[x + 1]] := TempChar;
Temp := SwapArray[x];
SwapArray[x] := SwapArray[x + 1];
SwapArray[x + 1] := Temp;
x := x + 2
end
end
end.
For those looking for less of a mathematical solution but utilizing data structure in JS. A combination of maps, set, and recursion:
function findUnmatchedCouples(people, couples) {
const copy = [...people]
if (copy.length == 0) {
return;
}
const p = copy[0];
copy.splice(0, 1);
for (let i = 0; i < copy.length; i++) {
const q = copy[i];
if (matchedThisRound.has(p)) {
continue;
}
if (!isCouple(p, q)) {
copy.splice(i, 1);
findUnmatchedCouples(copy, couples);
couples.push([p, q]);
matchedThisRound.add(p);
registerCouple(p, q)
}
}
return couples;
}
function isCouple(p, q) {
if (!registered.has(p)) {
return false;
} else {
const currentset = registered.get(p)
if (currentset.has(q)) {
//console.log(`${p} already matched with ${q}`)
return true;
}
currentset.add(q)
registered.set(p, currentset)
}
return false;
}
function registerCouple(p, q) {
if (!registered.has(p)) {
registered.set(p, new Set([q]))
} else {
const currentset = registered.get(p)
currentset.add(q)
registered.set(p, currentset)
}
}
//Start Secret Santa
const people = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
for (let i = 0; i < 10; i++) {
matchedThisRound = new Set([]);
console.log(`round ${i}`)
console.log(findUnmatchedCouples(people, []))
}
Let's say I have the following list: [2, 1, 4, 6, 3, 7]. I also have some method that sorts any list. However, I want to perform a sort across only elements at indices 1, 2, & 4, i.e. the sublist [1, 4, 3]. Sorting across this sublist produces [1, 3, 4]. How can get the original list such that I only sort across indices 1, 2, and 4, i.e., [2, 1, 3, 6, 4, 7]?
The easiest way is probably to use an extra level of indirection. For example, create a list (here meaning just some linear collection, not necessarily a linked list) of the indexes of the three elements you want to sort, and code to do comparison/swapping through that layer of indirection.
Thanks to the suggestion by Jerry Coffin, here's the solution in Java for those who are interested:
import java.util.List;
import java.util.AbstractList;
import java.util.Arrays;
public class ExtendedSubList<E> extends AbstractList<E>
{
protected final List<E> parent;
protected final int[] indices;
public static <E> List<E> subList(final List<E> parent, int ... indices)
{
if (parent == null)
throw new IllegalArgumentException("parent == null");
if (indices == null)
throw new IllegalArgumentException("indices == null");
for (int i = 0; i < indices.length; i++)
if (!(0 <= indices[i] && indices[i] < parent.size()))
throw new IllegalArgumentException(String.format("index %d (at position %d) is not in bounds", indices[i], i));
Arrays.sort(indices);
return new ExtendedSubList(parent, indices);
}
protected ExtendedSubList(List<E> parent, int[] indices)
{
this.parent = parent;
this.indices = indices;
}
public E get(int index)
{
return parent.get(indices[index]);
}
public int size()
{
return indices.length;
}
public E set(int index, E element)
{
return parent.set(indices[index], element);
}
}
Usage example:
List<Integer> list = Arrays.asList(2, 1, 4, 6, 3, 7);
Collections.sort(ExtendedSubList.subList(list), 1, 2, 4);
The resulting list would produce: [2, 1, 3, 6, 4, 7].
The following Python code does the job. It may differ from Jerry Coffins accepted answer as rather than sorting through indirection it extracts the values, sorts, then inserts them back.
data = [7, 6, 5, 4, 3, 2, 1, 0]
indices = sorted([1,2,4])
values = [data[i] for i in indices] # [6, 5, 3]
values.sort() # [3, 5, 6]
for index, value in zip(indices, values):
data[index] = value
print (data) # [7, 3, 5, 4, 6, 2, 1, 0]
The original indices should be
sorted for things to work.
The corresponding values are
extracted.
The values are sorted.
The for loop puts the sorted values
back into the original array.
I needed an algorithm to generate all possible partitions of a positive number, and I came up with one (posted as an answer), but it's exponential time.
The algorithm should return all the possible ways a number can be expressed as the sum of positive numbers less than or equal to itself. So for example for the number 5, the result would be:
5
4+1
3+2
3+1+1
2+2+1
2+1+1+1
1+1+1+1+1
So my question is: is there a more efficient algorithm for this?
EDIT: Question was titled "Sum decomposition of a number", since I didn't really know what this was called. ShreevatsaR pointed out that they were called "partitions," so I edited the question title accordingly.
It's called Partitions. [Also see Wikipedia: Partition (number theory).]
The number of partitions p(n) grows exponentially, so anything you do to generate all partitions will necessarily have to take exponential time.
That said, you can do better than what your code does. See this, or its updated version in Python Algorithms and Data Structures by David Eppstein.
Here's my solution (exponential time) in Python:
q = { 1: [[1]] }
def decompose(n):
try:
return q[n]
except:
pass
result = [[n]]
for i in range(1, n):
a = n-i
R = decompose(i)
for r in R:
if r[0] <= a:
result.append([a] + r)
q[n] = result
return result
>>> decompose(5)
[[5], [4, 1], [3, 2], [3, 1, 1], [2, 2, 1], [2, 1, 1, 1], [1, 1, 1, 1, 1]]
When you ask to more efficient algorithm, I don't know which to compare. But here is one algorithm written in straight forward way (Erlang):
-module(partitions).
-export([partitions/1]).
partitions(N) -> partitions(N, N).
partitions(N, Max) when N > 0 ->
[[X | P]
|| X <- lists:seq(min(N, Max), 1, -1),
P <- partitions(N - X, X)];
partitions(0, _) -> [[]];
partitions(_, _) -> [].
It is exponential in time (same as Can Berk Güder's solution in Python) and linear in stack space. But using same trick, memoization, you can achieve big improvement by save some memory and less exponent. (It's ten times faster for N=50)
mp(N) ->
lists:foreach(fun (X) -> put(X, undefined) end,
lists:seq(1, N)), % clean up process dictionary for sure
mp(N, N).
mp(N, Max) when N > 0 ->
case get(N) of
undefined -> R = mp(N, 1, Max, []), put(N, R), R;
[[Max | _] | _] = L -> L;
[[X | _] | _] = L ->
R = mp(N, X + 1, Max, L), put(N, R), R
end;
mp(0, _) -> [[]];
mp(_, _) -> [].
mp(_, X, Max, R) when X > Max -> R;
mp(N, X, Max, R) ->
mp(N, X + 1, Max, prepend(X, mp(N - X, X), R)).
prepend(_, [], R) -> R;
prepend(X, [H | T], R) -> prepend(X, T, [[X | H] | R]).
Anyway you should benchmark for your language and purposes.
Here's a much more long-winded way of doing it (this is what I did before I knew the term "partition", which enabled me to do a google search):
def magic_chunker (remainder, chunkSet, prevChunkSet, chunkSets):
if remainder > 0:
if prevChunkSet and (len(prevChunkSet) > len(chunkSet)): # counting down from previous
# make a chunk that is one less than relevant one in the prevChunkSet
position = len(chunkSet)
chunk = prevChunkSet[position] - 1
prevChunkSet = [] # clear prevChunkSet, no longer need to reference it
else: # begins a new countdown;
if chunkSet and (remainder > chunkSet[-1]): # no need to do iterations any greater than last chunk in this set
chunk = chunkSet[-1]
else: # i.e. remainder is less than or equal to last chunk in this set
chunk = remainder #else use the whole remainder for this chunk
chunkSet.append(chunk)
remainder -= chunk
magic_chunker(remainder, chunkSet, prevChunkSet, chunkSets)
else: #i.e. remainder==0
chunkSets.append(list(chunkSet)) #save completed partition
prevChunkSet = list(chunkSet)
if chunkSet[-1] > 1: # if the finalchunk was > 1, do further recursion
remainder = chunkSet.pop() #remove last member, and use it as remainder
magic_chunker(remainder, chunkSet, prevChunkSet, chunkSets)
else: # last chunk is 1
if chunkSet[0]==1: #the partition started with 1, we know we're finished
return chunkSets
else: #i.e. still more chunking to go
# clear back to last chunk greater than 1
while chunkSet[-1]==1:
remainder += chunkSet.pop()
remainder += chunkSet.pop()
magic_chunker(remainder, chunkSet, prevChunkSet, chunkSets)
partitions = []
magic_chunker(10, [], [], partitions)
print partitions
>> [[10], [9, 1], [8, 2], [8, 1, 1], [7, 3], [7, 2, 1], [7, 1, 1, 1], [6, 4], [6, 3, 1], [6, 2, 2], [6, 2, 1, 1], [6, 1, 1, 1, 1], [5, 5], [5, 4, 1], [5, 3, 2], [5, 3, 1, 1], [5, 2, 2, 1], [5, 2, 1, 1, 1], [5, 1, 1, 1, 1, 1], [4, 4, 2], [4, 4, 1, 1], [4, 3, 3], [4, 3, 2, 1], [4, 3, 1, 1, 1], [4, 2, 2, 2], [4, 2, 2, 1, 1], [4, 2, 1, 1, 1, 1], [4, 1, 1, 1, 1, 1, 1], [3, 3, 3, 1], [3, 3, 2, 2], [3, 3, 2, 1, 1], [3, 3, 1, 1, 1, 1], [3, 2, 2, 2, 1], [3, 2, 2, 1, 1, 1], [3, 2, 1, 1, 1, 1, 1], [3, 1, 1, 1, 1, 1, 1, 1], [2, 2, 2, 2, 2], [2, 2, 2, 2, 1, 1], [2, 2, 2, 1, 1, 1, 1], [2, 2, 1, 1, 1, 1, 1, 1], [2, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]]
Java implementation. Could benefit from memoization.
public class Partition {
/**
* partition returns a list of int[] that represent all distinct partitions of n.
*/
public static List<int[]> partition(int n) {
List<Integer> partial = new ArrayList<Integer>();
List<int[]> partitions = new ArrayList<int[]>();
partition(n, partial, partitions);
return partitions;
}
/**
* If n=0, it copies the partial solution into the list of complete solutions.
* Else, for all values i less than or equal to n, put i in the partial solution and partition the remainder n-i.
*/
private static void partition(int n, List<Integer> partial, List<int[]> partitions) {
//System.out.println("partition " + n + ", partial solution: " + partial);
if (n == 0) {
// Complete solution is held in 'partial' --> add it to list of solutions
partitions.add(toArray(partial));
} else {
// Iterate through all numbers i less than n.
// Avoid duplicate solutions by ensuring that the partial array is always non-increasing
for (int i=n; i>0; i--) {
if (partial.isEmpty() || partial.get(partial.size()-1) >= i) {
partial.add(i);
partition(n-i, partial, partitions);
partial.remove(partial.size()-1);
}
}
}
}
/**
* Helper method: creates a new integer array and copies the contents of the list into the array.
*/
private static int[] toArray(List<Integer> list) {
int i = 0;
int[] arr = new int[list.size()];
for (int val : list) {
arr[i++] = val;
}
return arr;
}
}
Here's a solution in using paramorphisms that I wrote in Haskell.
import Numeric.Natural (Natural)
import Control.Monad (join)
import Data.List (nub)
import Data.Functor.Foldable (ListF (..), para)
partitions :: Natural -> [[Natural]]
partitions = para algebra
where algebra Nothing = []
algebra (Just (0,_)) = [[1]]
algebra (Just (_, past)) = (nub . (getAll =<<)) (fmap (1:) past)
getAll :: [Natural] -> [[Natural]]
getAll = fmap (dropWhile (==0) . sort) . subsets
where subsets xs = flip sumIndicesAt xs <$> indices xs
indices :: [Natural] -> [[Natural]]
indices = join . para algebra
where algebra Nil = []
algebra (Cons x (xs, [])) = [[x:xs]]
algebra (Cons x (xs, past)) = (:) <$> [x:xs,[]] <*> past
It's definitely not the most efficient one around, but I think it's quite elegant and it's certainly instructive.
here is the java code for this question
static void printArray(int p[], int n){
for (int i = 0; i < n; i++)
System.out.print(p[i]+" ");
System.out.println();
}
// Function to generate all unique partitions of an integer
static void printAllUniqueParts(int n) {
int[] p = new int[n]; // An array to store a partition
int k = 0; // Index of last element in a partition
p[k] = n; // Initialize first partition as number itself
// This loop first prints current partition, then generates next
// partition. The loop stops when the current partition has all 1s
while (true) {
// print current partition
printArray(p, k + 1);
// Generate next partition
// Find the rightmost non-one value in p[]. Also, update the
// rem_val so that we know how much value can be accommodated
int rem_val = 0;
while (k >= 0 && p[k] == 1) {
rem_val += p[k];
k--;
}
// if k < 0, all the values are 1 so there are no more partitions
if (k < 0){
break;
}
// Decrease the p[k] found above and adjust the rem_val
p[k]--;
rem_val++;
while (rem_val > p[k]) {
p[k + 1] = p[k];
rem_val = rem_val - p[k];
k++;
}
p[k + 1] = rem_val;
k++;
}
}
public static void main(String[] args) {
System.out.println("All Unique Partitions of 5");
printAllUniqueParts(5);
System.out.println("All Unique Partitions of 7");
printAllUniqueParts(7);
System.out.println("All Unique Partitions of 9");
printAllUniqueParts(8);
}
Another Java solution. It starts by creating first partition which is only the given number. Then it goes in while loop which is finding the last number in last created partition which is bigger then 1. From that number it moves 1 to next number in array. If next number would end up being the same as the found number it moves to the next in line. Loop stops when first number of last created partition is 1. This works because at all times numbers in all partitions are sorted in descending order.
Example with number 5. First it creates first partition which is just number 5. Then it finds last number in last partition that is greater then 1. Since our last partition is array [5, 0, 0, 0, 0] it founds number 5 at index 0. Then it takes one from 5 and moves it to next position. That is how we get partition [4, 1, 0, 0, 0]. It goes into the loop again. Now it takes one from 4 and moves it up so we get [3, 2, 0, 0, 0]. Then the same thing and we get [3, 1, 1, 0, 0]. On next iteration we get [2, 2, 1, 0, 0]. Now it takes one from second 2 and tries to move it to index 2 where we have 1. It will skip to the next index because we would also get 2 and we would have partition [2, 1, 2, 0, 0] which is just duplicate of the last one. instead we get [2, 1, 1, 1, 0]. And in the last step we get to [1, 1, 1, 1, 1] and loop exists since first number of new partition is 1.
private static List<int[]> getNumberPartitions(int n) {
ArrayList<int[]> result = new ArrayList<>();
int[] initial = new int[n];
initial[0] = n;
result.add(initial);
while (result.get(result.size() - 1)[0] > 1) {
int[] lastPartition = result.get(result.size() - 1);
int posOfLastNotOne = 0;
for(int k = lastPartition.length - 1; k >= 0; k--) {
if (lastPartition[k] > 1) {
posOfLastNotOne = k;
break;
}
}
int[] newPartition = new int[n];
for (int j = posOfLastNotOne + 1; j < lastPartition.length; j++) {
if (lastPartition[posOfLastNotOne] - 1 > lastPartition[j]) {
System.arraycopy(lastPartition, 0, newPartition, 0, lastPartition.length);
newPartition[posOfLastNotOne]--;
newPartition[j]++;
result.add(newPartition);
break;
}
}
}
return result;
}
Here is my Rust implementation (inspired by Python Algorithms and Data Structures):
#[derive(Clone)]
struct PartitionIter {
pub n: u32,
partition: Vec<u32>,
last_not_one_index: usize,
started: bool,
finished: bool
}
impl PartitionIter {
pub fn new(n: u32) -> PartitionIter {
PartitionIter {
n,
partition: Vec::with_capacity(n as usize),
last_not_one_index: 0,
started: false,
finished: false,
}
}
}
impl Iterator for PartitionIter {
type Item = Vec<u32>;
fn next(&mut self) -> Option<Self::Item> {
if self.finished {
return None
}
if !self.started {
self.partition.push(self.n);
self.started = true;
return Some(self.partition.clone());
} else if self.n == 1 {
return None;
}
if self.partition[self.last_not_one_index] == 2 {
self.partition[self.last_not_one_index] = 1;
self.partition.push(1);
if self.last_not_one_index == 0 {
self.finished = true;
} else {
self.last_not_one_index -= 1;
}
return Some(self.partition.clone())
}
let replacement = self.partition[self.last_not_one_index] - 1;
let total_replaced = replacement + (self.partition.len() - self.last_not_one_index) as u32;
let reps = total_replaced / replacement;
let rest = total_replaced % replacement;
self.partition.drain(self.last_not_one_index..);
self.partition.extend_from_slice(&vec![replacement; reps as usize]);
if rest > 0 {
self.partition.push(rest);
}
self.last_not_one_index = self.partition.len() - (self.partition.last().cloned().unwrap() == 1) as usize - 1;
Some(self.partition.clone())
}
}