For example, string "AAABBB" will have permutations:
"ABAABB",
"BBAABA",
"ABABAB",
etc
What's a good algorithm for generating the permutations? (And what's its time complexity?)
For a multiset, you can solve recursively by position (JavaScript code):
function f(multiset,counters,result){
if (counters.every(x => x === 0)){
console.log(result);
return;
}
for (var i=0; i<counters.length; i++){
if (counters[i] > 0){
_counters = counters.slice();
_counters[i]--;
f(multiset,_counters,result + multiset[i]);
}
}
}
f(['A','B'],[3,3],'');
This is not full answer, just an idea.
If your strings has fixed number of only two letters I'll go with binary tree and good recursion function.
Each node is object that contains name with prefix of parent name and suffix A or B furthermore it have numbers of A and B letters in the name.
Node constructor gets name of parent and number of A and B from parent so it needs only to add 1 to number of A or B and one letter to name.
It doesn't construct next node if there is more than three A (in case of A node) or B respectively, or their sum is equal to the length of starting string.
Now you can collect leafs of 2 trees (their names) and have all permutations that you need.
Scala or some functional language (with object-like features) would be perfect for implementing this algorithm. Hope this helps or just sparks some ideas.
Since you actually want to generate the permutations instead of just counting them, the best complexity you can hope for is O(size_of_output).
Here's a good solution in java that meets that bound and runs very quickly, while consuming negligible space. It first sorts the letters to find the lexographically smallest permutation, and then generates all permutations in lexographic order.
It's known as the Pandita algorithm: https://en.wikipedia.org/wiki/Permutation#Generation_in_lexicographic_order
import java.util.Arrays;
import java.util.function.Consumer;
public class UniquePermutations
{
static void generateUniquePermutations(String s, Consumer<String> consumer)
{
char[] array = s.toCharArray();
Arrays.sort(array);
for (;;)
{
consumer.accept(String.valueOf(array));
int changePos=array.length-2;
while (changePos>=0 && array[changePos]>=array[changePos+1])
--changePos;
if (changePos<0)
break; //all done
int swapPos=changePos+1;
while(swapPos+1 < array.length && array[swapPos+1]>array[changePos])
++swapPos;
char t = array[changePos];
array[changePos] = array[swapPos];
array[swapPos] = t;
for (int i=changePos+1, j = array.length-1; i < j; ++i,--j)
{
t = array[i];
array[i] = array[j];
array[j] = t;
}
}
}
public static void main (String[] args) throws java.lang.Exception
{
StringBuilder line = new StringBuilder();
generateUniquePermutations("banana", s->{
if (line.length() > 0)
{
if (line.length() + s.length() >= 75)
{
System.out.println(line.toString());
line.setLength(0);
}
else
line.append(" ");
}
line.append(s);
});
System.out.println(line);
}
}
Here is the output:
aaabnn aaanbn aaannb aabann aabnan aabnna aanabn aananb aanban aanbna
aannab aannba abaann abanan abanna abnaan abnana abnnaa anaabn anaanb
anaban anabna ananab ananba anbaan anbana anbnaa annaab annaba annbaa
baaann baanan baanna banaan banana bannaa bnaaan bnaana bnanaa bnnaaa
naaabn naaanb naaban naabna naanab naanba nabaan nabana nabnaa nanaab
nanaba nanbaa nbaaan nbaana nbanaa nbnaaa nnaaab nnaaba nnabaa nnbaaa
We have two strings a and b respectively. The length of a is greater than or equal to b. We have to find out the longest common substring. If there are multiple answers then we have to output the substring which comes earlier in b (earlier as in whose starting index comes first).
Note: The length of a and b can be up to 106.
I tried to find the longest common substring using suffix array (sorting the suffixes using quicksort). For the case when there is more than one answer, I tried pushing all the common substrings in a stack which are equal to the length of the longest common substring.
I wanted to know is there any faster way to do so?
Build a suffix tree of a string a$b, that is, a concatenated with some character like $ not occurring in both strings, then concatenated with b. A (compressed) suffix tree can be built in O(|a|+|b|) time and memory, and have O(|a|+|b|) nodes.
Now, for each node, we know its depth (the length of the string obtained by starting from the root and traversing the tree down to that node). We also can keep track of two boolean quantities: whether this node was visited during the build phase corresponding to a, and whether it was visited during the build phase corresponding to b (for example, we might as well build the two trees separately and then merge them using pre-order traversal). Now, the task boils down to finding the deepest vertex which was visited during both phases, which can be done by a single pre-order traversal. The case of multiple answers should be easy to handle.
This Wikipedia page contains another (brief) overview of the technique.
This is longest substring,what you are looking for is it with repetition or without .
please go through this it might be helpful.
http://www.programcreek.com/2013/02/leetcode-longest-substring-without-repeating-characters-java/
import java.util.Scanner;
public class JavaApplication8 {
public static int find(String s1,String s2){
int n = s1.length();
int m = s2.length();
int ans = 0;
int[] a = new int[m];
int b[] = new int[m];
for(int i = 0;i<n;i++){
for(int j = 0;j<m;j++){
if(s1.charAt(i)==s2.charAt(j)){
if(i==0 || j==0 )a[j] = 1;
else{
a[j] = b[j-1] + 1;
}
ans = Math.max(ans, a[j]);
}
}
int[] c = a;
a = b;
b = c;
}
return ans;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String s1 = sc.next();
String s2 = sc.next();
System.out.println(find(s1,s2));
}
}
Time Complexity O(N)
Space Complexity O(N)
package main
import (
"fmt"
"strings"
)
func main(){
fmt.Println(lcs("CLCL","LCLC"))
}
func lcs(s1,s2 string)(max int,str string){
str1 := strings.Split(s1,"")
str2 := strings.Split(s2,"")
fmt.Println(str1,str2)
str = ""
mnMatrix := [4][4]int{}
for i:=0;i<len(str1);i++{
for j:=0;j<len(str2);j++{
if str1[i]==str2[j]{
if i==0 || j==0 {
mnMatrix[i][j] = 1
max = 1
//str = str1[i]
}else{
mnMatrix[i][j] = mnMatrix[i-1][j-1]+1
max = mnMatrix[i][j]
str = ""
for k:=max;k>=1;k--{
str = str + str2[k]
//fmt.Println(str)
}
}
}else{
mnMatrix[i][j] = 0
}
}
}
fmt.Println(mnMatrix)
return max, str
}
enter code here
I'm looking for a good (easy to implement, intuitive, etc.) recursive method of generating all binary strings of length n, where 1 <= n <= 35.
I would appreciate ideas for a pseudo-code algorithm (no language-specific tricks).
LE: okay, I did go overboard with the upper limit. My intention was to avoid solutions that use the binary representation of a counter from 1 to 1 << n.
Here's an example of recursion in C++.
vector<string> answer;
void getStrings( string s, int digitsLeft )
{
if( digitsLeft == 0 ) // the length of string is n
answer.push_back( s );
else
{
getStrings( s + "0", digitsLeft - 1 );
getStrings( s + "1", digitsLeft - 1 );
}
}
getStrings( "", n ); // initial call
According to the Divide et Impera paradigm, the problem of generating all binary strings of length n can be splitted in two subproblems: the problem of printing all binary strings of lenght n-1 preceeded by a 0 and the one of printing all binary strings of lenght n-1 preceeded by a 1. So the following pseudocode solves the problem:
generateBinary(length, string)
if(length > 0)
generateBinary(length-1, string + "0")
generateBinary(length-1, string + "1")
else
print(string)
def genBins(n):
"""
generate all the binary strings with n-length
"""
max_int = '0b' + '1' * n
for i in range(0, int(max_int, 2)+1, 1):
yield str(format(i, 'b').zfill(n))
if __name__ == '__main__':
print(list(genBins(5)))
The problem you have can be solved with a Backtracking algorithm.
Pseudo-code for such an algorithm is:
fun(input, n)
if( base_case(input, n) )
//print result
else
//choose from pool of choices
//explorer the rest of choices from what's left
//unchoose
Implementation:
Base case: we want to print our result string when its size is equal to n
Recursive case:
our pool of choices consists of 0 and 1
choosing in this case means take 0 or 1 and add it to the input as last character
explore by recursing, where we pass the new input value from the choose step until base case is reached
un-choosing in this case means remove the last character
function binary(n) {
binaryHelper('', n);
}
function binaryHelper(str, n) {
if (str.length === n) {
//base case
console.log(str); //print string
} else {
for (let bit = 0; bit < 2; bit++) {
str = str + bit; // choose
binaryHelper(str, n); // explore
str = str.slice(0, -1); // un-choose
}
}
}
console.log('Size 2 binary strings:');
binary(2);
console.log('Size 3 binary strings:');
binary(3);
You can re-write the code above like this, where you choose & un-choose by stateless transition from one loop iteration to another. This is less intuitive though.
function binary(n) {
binaryHelper('', n);
}
function binaryHelper(str, n) {
if(str.length === n) {
console.log(str);
} else {
for(let bit = 0; bit < 2; bit++) {
binaryHelper(str+bit, n);
}
}
}
console.log('Size 2 binary strings:');
binary(2);
console.log('Size 3 binary strings:');
binary(3);
How do you print numbers of form 2^i * 5^j in increasing order.
For eg:
1, 2, 4, 5, 8, 10, 16, 20
This is actually a very interesting question, especially if you don't want this to be N^2 or NlogN complexity.
What I would do is the following:
Define a data structure containing 2 values (i and j) and the result of the formula.
Define a collection (e.g. std::vector) containing this data structures
Initialize the collection with the value (0,0) (the result is 1 in this case)
Now in a loop do the following:
Look in the collection and take the instance with the smallest value
Remove it from the collection
Print this out
Create 2 new instances based on the instance you just processed
In the first instance increment i
In the second instance increment j
Add both instances to the collection (if they aren't in the collection yet)
Loop until you had enough of it
The performance can be easily tweaked by choosing the right data structure and collection.
E.g. in C++, you could use an std::map, where the key is the result of the formula, and the value is the pair (i,j). Taking the smallest value is then just taking the first instance in the map (*map.begin()).
I quickly wrote the following application to illustrate it (it works!, but contains no further comments, sorry):
#include <math.h>
#include <map>
#include <iostream>
typedef __int64 Integer;
typedef std::pair<Integer,Integer> MyPair;
typedef std::map<Integer,MyPair> MyMap;
Integer result(const MyPair &myPair)
{
return pow((double)2,(double)myPair.first) * pow((double)5,(double)myPair.second);
}
int main()
{
MyMap myMap;
MyPair firstValue(0,0);
myMap[result(firstValue)] = firstValue;
while (true)
{
auto it=myMap.begin();
if (it->first < 0) break; // overflow
MyPair myPair = it->second;
std::cout << it->first << "= 2^" << myPair.first << "*5^" << myPair.second << std::endl;
myMap.erase(it);
MyPair pair1 = myPair;
++pair1.first;
myMap[result(pair1)] = pair1;
MyPair pair2 = myPair;
++pair2.second;
myMap[result(pair2)] = pair2;
}
}
This is well suited to a functional programming style. In F#:
let min (a,b)= if(a<b)then a else b;;
type stream (current, next)=
member this.current = current
member this.next():stream = next();;
let rec merge(a:stream,b:stream)=
if(a.current<b.current) then new stream(a.current, fun()->merge(a.next(),b))
else new stream(b.current, fun()->merge(a,b.next()));;
let rec Squares(start) = new stream(start,fun()->Squares(start*2));;
let rec AllPowers(start) = new stream(start,fun()->merge(Squares(start*2),AllPowers(start*5)));;
let Results = AllPowers(1);;
Works well with Results then being a stream type with current value and a next method.
Walking through it:
I define min for completenes.
I define a stream type to have a current value and a method to return a new string, essentially head and tail of a stream of numbers.
I define the function merge, which takes the smaller of the current values of two streams and then increments that stream. It then recurses to provide the rest of the stream. Essentially, given two streams which are in order, it will produce a new stream which is in order.
I define squares to be a stream increasing in powers of 2.
AllPowers takes the start value and merges the stream resulting from all squares at this number of powers of 5. it with the stream resulting from multiplying it by 5, since these are your only two options. You effectively are left with a tree of results
The result is merging more and more streams, so you merge the following streams
1, 2, 4, 8, 16, 32...
5, 10, 20, 40, 80, 160...
25, 50, 100, 200, 400...
.
.
.
Merging all of these turns out to be fairly efficient with tail recursio and compiler optimisations etc.
These could be printed to the console like this:
let rec PrintAll(s:stream)=
if (s.current > 0) then
do System.Console.WriteLine(s.current)
PrintAll(s.next());;
PrintAll(Results);
let v = System.Console.ReadLine();
Similar things could be done in any language which allows for recursion and passing functions as values (it's only a little more complex if you can't pass functions as variables).
For an O(N) solution, you can use a list of numbers found so far and two indexes: one representing the next number to be multiplied by 2, and the other the next number to be multiplied by 5. Then in each iteration you have two candidate values to choose the smaller one from.
In Python:
numbers = [1]
next_2 = 0
next_5 = 0
for i in xrange(100):
mult_2 = numbers[next_2]*2
mult_5 = numbers[next_5]*5
if mult_2 < mult_5:
next = mult_2
next_2 += 1
else:
next = mult_5
next_5 += 1
# The comparison here is to avoid appending duplicates
if next > numbers[-1]:
numbers.append(next)
print numbers
So we have two loops, one incrementing i and second one incrementing j starting both from zero, right? (multiply symbol is confusing in the title of the question)
You can do something very straightforward:
Add all items in an array
Sort the array
Or you need an other solution with more math analysys?
EDIT: More smart solution by leveraging similarity with Merge Sort problem
If we imagine infinite set of numbers of 2^i and 5^j as two independent streams/lists this problem looks very the same as well known Merge Sort problem.
So solution steps are:
Get two numbers one from the each of streams (of 2 and of 5)
Compare
Return smallest
get next number from the stream of the previously returned smallest
and that's it! ;)
PS: Complexity of Merge Sort always is O(n*log(n))
I visualize this problem as a matrix M where M(i,j) = 2^i * 5^j. This means that both the rows and columns are increasing.
Think about drawing a line through the entries in increasing order, clearly beginning at entry (1,1). As you visit entries, the row and column increasing conditions ensure that the shape formed by those cells will always be an integer partition (in English notation). Keep track of this partition (mu = (m1, m2, m3, ...) where mi is the number of smaller entries in row i -- hence m1 >= m2 >= ...). Then the only entries that you need to compare are those entries which can be added to the partition.
Here's a crude example. Suppose you've visited all the xs (mu = (5,3,3,1)), then you need only check the #s:
x x x x x #
x x x #
x x x
x #
#
Therefore the number of checks is the number of addable cells (equivalently the number of ways to go up in Bruhat order if you're of a mind to think in terms of posets).
Given a partition mu, it's easy to determine what the addable states are. Image an infinite string of 0s following the last positive entry. Then you can increase mi by 1 if and only if m(i-1) > mi.
Back to the example, for mu = (5,3,3,1) we can increase m1 (6,3,3,1) or m2 (5,4,3,1) or m4 (5,3,3,2) or m5 (5,3,3,1,1).
The solution to the problem then finds the correct sequence of partitions (saturated chain). In pseudocode:
mu = [1,0,0,...,0];
while (/* some terminate condition or go on forever */) {
minNext = 0;
nextCell = [];
// look through all addable cells
for (int i=0; i<mu.length; ++i) {
if (i==0 or mu[i-1]>mu[i]) {
// check for new minimum value
if (minNext == 0 or 2^i * 5^(mu[i]+1) < minNext) {
nextCell = i;
minNext = 2^i * 5^(mu[i]+1)
}
}
}
// print next largest entry and update mu
print(minNext);
mu[i]++;
}
I wrote this in Maple stopping after 12 iterations:
1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50
and the outputted sequence of cells added and got this:
1 2 3 5 7 10
4 6 8 11
9 12
corresponding to this matrix representation:
1, 2, 4, 8, 16, 32...
5, 10, 20, 40, 80, 160...
25, 50, 100, 200, 400...
First of all, (as others mentioned already) this question is very vague!!!
Nevertheless, I am going to give a shot based on your vague equation and the pattern as your expected result. So I am not sure the following will be true for what you are trying to do, however it may give you some idea about java collections!
import java.util.List;
import java.util.ArrayList;
import java.util.SortedSet;
import java.util.TreeSet;
public class IncreasingNumbers {
private static List<Integer> findIncreasingNumbers(int maxIteration) {
SortedSet<Integer> numbers = new TreeSet<Integer>();
SortedSet<Integer> numbers2 = new TreeSet<Integer>();
for (int i=0;i < maxIteration;i++) {
int n1 = (int)Math.pow(2, i);
numbers.add(n1);
for (int j=0;j < maxIteration;j++) {
int n2 = (int)Math.pow(5, i);
numbers.add(n2);
for (Integer n: numbers) {
int n3 = n*n1;
numbers2.add(n3);
}
}
}
numbers.addAll(numbers2);
return new ArrayList<Integer>(numbers);
}
/**
* Based on the following fuzzy question # StackOverflow
* http://stackoverflow.com/questions/7571934/printing-numbers-of-the-form-2i-5j-in-increasing-order
*
*
* Result:
* 1 2 4 5 8 10 16 20 25 32 40 64 80 100 125 128 200 256 400 625 1000 2000 10000
*/
public static void main(String[] args) {
List<Integer> numbers = findIncreasingNumbers(5);
for (Integer i: numbers) {
System.out.print(i + " ");
}
}
}
If you can do it in O(nlogn), here's a simple solution:
Get an empty min-heap
Put 1 in the heap
while (you want to continue)
Get num from heap
print num
put num*2 and num*5 in the heap
There you have it. By min-heap, I mean min-heap
As a mathematician the first thing I always think about when looking at something like this is "will logarithms help?".
In this case it might.
If our series A is increasing then the series log(A) is also increasing. Since all terms of A are of the form 2^i.5^j then all members of the series log(A) are of the form i.log(2) + j.log(5)
We can then look at the series log(A)/log(2) which is also increasing and its elements are of the form i+j.(log(5)/log(2))
If we work out the i and j that generates the full ordered list for this last series (call it B) then that i and j will also generate the series A correctly.
This is just changing the nature of the problem but hopefully to one where it becomes easier to solve. At each step you can either increase i and decrease j or vice versa.
Looking at a few of the early changes you can make (which I will possibly refer to as transforms of i,j or just transorms) gives us some clues of where we are going.
Clearly increasing i by 1 will increase B by 1. However, given that log(5)/log(2) is approx 2.3 then increasing j by 1 while decreasing i by 2 will given an increase of just 0.3 . The problem then is at each stage finding the minimum possible increase in B for changes of i and j.
To do this I just kept a record as I increased of the most efficient transforms of i and j (ie what to add and subtract from each) to get the smallest possible increase in the series. Then applied whichever one was valid (ie making sure i and j don't go negative).
Since at each stage you can either decrease i or decrease j there are effectively two classes of transforms that can be checked individually. A new transform doesn't have to have the best overall score to be included in our future checks, just better than any other in its class.
To test my thougths I wrote a sort of program in LinqPad. Key things to note are that the Dump() method just outputs the object to screen and that the syntax/structure isn't valid for a real c# file. Converting it if you want to run it should be easy though.
Hopefully anything not explicitly explained will be understandable from the code.
void Main()
{
double C = Math.Log(5)/Math.Log(2);
int i = 0;
int j = 0;
int maxi = i;
int maxj = j;
List<int> outputList = new List<int>();
List<Transform> transforms = new List<Transform>();
outputList.Add(1);
while (outputList.Count<500)
{
Transform tr;
if (i==maxi)
{
//We haven't considered i this big before. Lets see if we can find an efficient transform by getting this many i and taking away some j.
maxi++;
tr = new Transform(maxi, (int)(-(maxi-maxi%C)/C), maxi%C);
AddIfWorthwhile(transforms, tr);
}
if (j==maxj)
{
//We haven't considered j this big before. Lets see if we can find an efficient transform by getting this many j and taking away some i.
maxj++;
tr = new Transform((int)(-(maxj*C)), maxj, (maxj*C)%1);
AddIfWorthwhile(transforms, tr);
}
//We have a set of transforms. We first find ones that are valid then order them by score and take the first (smallest) one.
Transform bestTransform = transforms.Where(x=>x.I>=-i && x.J >=-j).OrderBy(x=>x.Score).First();
//Apply transform
i+=bestTransform.I;
j+=bestTransform.J;
//output the next number in out list.
int value = GetValue(i,j);
//This line just gets it to stop when it overflows. I would have expected an exception but maybe LinqPad does magic with them?
if (value<0) break;
outputList.Add(value);
}
outputList.Dump();
}
public int GetValue(int i, int j)
{
return (int)(Math.Pow(2,i)*Math.Pow(5,j));
}
public void AddIfWorthwhile(List<Transform> list, Transform tr)
{
if (list.Where(x=>(x.Score<tr.Score && x.IncreaseI == tr.IncreaseI)).Count()==0)
{
list.Add(tr);
}
}
// Define other methods and classes here
public class Transform
{
public int I;
public int J;
public double Score;
public bool IncreaseI
{
get {return I>0;}
}
public Transform(int i, int j, double score)
{
I=i;
J=j;
Score=score;
}
}
I've not bothered looking at the efficiency of this but I strongly suspect its better than some other solutions because at each stage all I need to do is check my set of transforms - working out how many of these there are compared to "n" is non-trivial. It is clearly related since the further you go the more transforms there are but the number of new transforms becomes vanishingly small at higher numbers so maybe its just O(1). This O stuff always confused me though. ;-)
One advantage over other solutions is that it allows you to calculate i,j without needing to calculate the product allowing me to work out what the sequence would be without needing to calculate the actual number itself.
For what its worth after the first 230 nunmbers (when int runs out of space) I had 9 transforms to check each time. And given its only my total that overflowed I ran if for the first million results and got to i=5191 and j=354. The number of transforms was 23. The size of this number in the list is approximately 10^1810. Runtime to get to this level was approx 5 seconds.
P.S. If you like this answer please feel free to tell your friends since I spent ages on this and a few +1s would be nice compensation. Or in fact just comment to tell me what you think. :)
I'm sure everyone one's might have got the answer by now, but just wanted to give a direction to this solution..
It's a Ctrl C + Ctrl V from
http://www.careercup.com/question?id=16378662
void print(int N)
{
int arr[N];
arr[0] = 1;
int i = 0, j = 0, k = 1;
int numJ, numI;
int num;
for(int count = 1; count < N; )
{
numI = arr[i] * 2;
numJ = arr[j] * 5;
if(numI < numJ)
{
num = numI;
i++;
}
else
{
num = numJ;
j++;
}
if(num > arr[k-1])
{
arr[k] = num;
k++;
count++;
}
}
for(int counter = 0; counter < N; counter++)
{
printf("%d ", arr[counter]);
}
}
The question as put to me was to return an infinite set of solutions. I pondered the use of trees, but felt there was a problem with figuring out when to harvest and prune the tree, given an infinite number of values for i & j. I realized that a sieve algorithm could be used. Starting from zero, determine whether each positive integer had values for i and j. This was facilitated by turning answer = (2^i)*(2^j) around and solving for i instead. That gave me i = log2 (answer/ (5^j)). Here is the code:
class Program
{
static void Main(string[] args)
{
var startTime = DateTime.Now;
int potential = 0;
do
{
if (ExistsIandJ(potential))
Console.WriteLine("{0}", potential);
potential++;
} while (potential < 100000);
Console.WriteLine("Took {0} seconds", DateTime.Now.Subtract(startTime).TotalSeconds);
}
private static bool ExistsIandJ(int potential)
{
// potential = (2^i)*(5^j)
// 1 = (2^i)*(5^j)/potential
// 1/(2^1) = (5^j)/potential or (2^i) = potential / (5^j)
// i = log2 (potential / (5^j))
for (var j = 0; Math.Pow(5,j) <= potential; j++)
{
var i = Math.Log(potential / Math.Pow(5, j), 2);
if (i == Math.Truncate(i))
return true;
}
return false;
}
}