# Longest common subsequence with permutations allowed

Given two strings in lowercase, find the longest string whose permutations are subsequences of given two strings. The output longest string must be sorted.

Examples:

```Input  :  str1 = "pink", str2 = "kite"
Output : "ik"
The string "ik" is the longest sorted string
whose one permutation "ik" is subsequence of
"pink" and another permutation "ki" is
subsequence of "kite".

Input  : str1 = "working", str2 = "women"
Output : "now"

Input  : str1 = "geeks" , str2 = "cake"
Output : "ek"

Input  : str1 = "aaaa" , str2 = "baba"
Output : "aa"
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to count characters in both strings.

1. calculate frequency of characters for each string and store them in their respective count arrays, say count1[] for str1 and count2[] for str2.
2. Now we have count arrays for 26 characters. So traverse count1[] and for any index ‘i’ append character (‘a’+i) in resultant string ‘result’ by min(count1[i], count2[i]) times.
3. Since we traverse count array in ascending order, our final string characters will be in sorted order.
4. ```// C++ program to find LCS with permutations allowed
#include<bits/stdc++.h>
using namespace std;

// Function to calculate longest string
// str1     --> first string
// str2     --> second string
// count1[]  --> hash array to calculate frequency
//              of characters in str1
// count[2]  --> hash array to calculate frequency
//              of characters in str2
// result   --> resultant longest string whose
// permutations are sub-sequence of given two strings
void longestString(string str1, string str2)
{
int count1[26] = {0}, count2[26]= {0};

// calculate frequency  of characters
for (int i=0; i<str1.length(); i++)
count1[str1[i]-'a']++;
for (int i=0; i<str2.length(); i++)
count2[str2[i]-'a']++;

// Now traverse hash array
string result;
for (int i=0; i<26; i++)

// append character ('a'+i) in resultant
// string 'result' by min(count1[i],count2i])
// times
for (int j=1; j<=min(count1[i],count2[i]); j++)
result.push_back('a' + i);

cout << result;
}

// Driver program to run the case
int main()
{
string str1 = "geeks", str2 = "cake";
longestString(str1, str2);
return 0;
}
```

Output:

```ek
```

Time Complexity : O(m + n) where m and n are lengths of input strings.
Auxiliary Space : O(1)

If you have another approach to solve this problem then please share.

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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