# 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++

 `// C++ program to find LCS with permutations allowed ` `#include ` `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  --> 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 = {0}, count2= {0}; ` ` `  `    ``// calculate frequency  of characters ` `    ``for` `(``int` `i=0; i

## Java

 `//Java program to find LCS with permutations allowed ` ` `  `class` `GFG { ` ` `  `// Function to calculate longest String  ` `// str1     --> first String  ` `// str2     --> second String  ` `// count1[] --> hash array to calculate frequency  ` `//             of characters in str1  ` `// count --> hash array to calculate frequency  ` `//             of characters in str2  ` `// result --> resultant longest String whose  ` `// permutations are sub-sequence of given two strings  ` `    ``static` `void` `longestString(String str1, String str2) { ` `        ``int` `count1[] = ``new` `int``[``26``], count2[] = ``new` `int``[``26``]; ` ` `  `        ``// calculate frequency of characters  ` `        ``for` `(``int` `i = ``0``; i < str1.length(); i++) { ` `            ``count1[str1.charAt(i) - ``'a'``]++; ` `        ``} ` `        ``for` `(``int` `i = ``0``; i < str2.length(); i++) { ` `            ``count2[str2.charAt(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 <= Math.min(count1[i], count2[i]); j++) { ` `                ``result += (``char``)(``'a'` `+ i); ` `            ``} ` `        ``} ` ` `  `        ``System.out.println(result); ` `    ``} ` ` `  `// Driver program to run the case  ` `    ``public` `static` `void` `main(String[] args) { ` `        ``String str1 = ``"geeks"``, str2 = ``"cake"``; ` `        ``longestString(str1, str2); ` ` `  `    ``} ` `} ` `/* This java code is contributed by 29AjayKumar*/`

## Python 3

 `# Python 3 program to find LCS ` `# with permutations allowed ` ` `  `# Function to calculate longest string ` `# str1     --> first string ` `# str2     --> second string ` `# count1[] --> hash array to calculate frequency ` `#               of characters in str1 ` `# count --> hash array to calculate frequency ` `#               of characters in str2 ` `# result --> resultant longest string whose ` `# permutations are sub-sequence  ` `# of given two strings ` `def` `longestString(str1, str2): ` ` `  `    ``count1 ``=` `[``0``] ``*` `26` `    ``count2 ``=` `[``0``] ``*` `26` ` `  `    ``# calculate frequency of characters ` `    ``for` `i ``in` `range``( ``len``(str1)): ` `        ``count1[``ord``(str1[i]) ``-` `ord``(``'a'``)] ``+``=` `1` `    ``for` `i ``in` `range``(``len``(str2)): ` `        ``count2[``ord``(str2[i]) ``-` `ord``(``'a'``)] ``+``=` `1` ` `  `    ``# Now traverse hash array ` `    ``result ``=` `"" ` `    ``for` `i ``in` `range``(``26``): ` ` `  `        ``# append character ('a'+i) in  ` `        ``# resultant string 'result' by  ` `        ``# min(count1[i],count2i]) times ` `        ``for` `j ``in` `range``(``1``, ``min``(count1[i], ` `                              ``count2[i]) ``+` `1``): ` `            ``result ``=` `result ``+` `chr``(``ord``(``'a'``) ``+` `i) ` ` `  `    ``print``(result) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `     `  `    ``str1 ``=` `"geeks"` `    ``str2 ``=` `"cake"` `    ``longestString(str1, str2) ` ` `  `# This code is contributed by ita_c `

## C#

 `// C# program to find LCS with  ` `// permutations allowed ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to calculate longest String  ` `// str1 --> first String  ` `// str2 --> second String  ` `// count1[] --> hash array to calculate   ` `//        frequency of characters in str1  ` `// count --> hash array to calculate  ` `//         frequency of characters in str2  ` `// result --> resultant longest String whose  ` `// permutations are sub-sequence of ` `// given two strings  ` `static` `void` `longestString(String str1,  ` `                          ``String str2)  ` `{ ` `    ``int` `[]count1 = ``new` `int``; ` `    ``int` `[]count2 = ``new` `int``; ` ` `  `    ``// 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 <= Math.Min(count1[i],  ` `                               ``count2[i]); j++) ` `        ``{ ` `            ``result += (``char``)(``'a'` `+ i); ` `        ``} ` `    ``} ` ` `  `Console.Write(result); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``String str1 = ``"geeks"``, str2 = ``"cake"``; ` `    ``longestString(str1, str2); ` `} ` `} ` ` `  `// This code is contributed ` `// by PrinciRaj1992 `

## PHP

 ` first string ` `// str2     --> second string ` `// count1[] --> hash array to calculate frequency ` `//                of characters in str1 ` `// count --> hash array to calculate frequency ` `//                of characters in str2 ` `// result --> resultant longest string whose ` `// permutations are sub-sequence of given two strings ` `function` `longestString(``\$str1``, ``\$str2``) ` `{ ` `    ``\$count1` `= ``array_fill``(0, 26, NULL); ` `    ``\$count2` `= ``array_fill``(0, 26, NULL); ` ` `  `    ``// calculate frequency of characters ` `    ``for` `(``\$i` `= 0; ``\$i` `< ``strlen``(``\$str1``); ``\$i``++) ` `        ``\$count1``[ord(``\$str1``[``\$i``]) - ord(``'a'``)]++; ` `    ``for` `(``\$i` `= 0; ``\$i` `< ``strlen``(``\$str2``); ``\$i``++) ` `        ``\$count2``[ord(``\$str2``[``\$i``]) - ord(``'a'``)]++; ` ` `  `    ``// Now traverse hash array ` `    ``\$result` `= ``""``; ` `    ``for` `(``\$i` `= 0; ``\$i` `< 26; ``\$i``++) ` ` `  `        ``// append character ('a'+i) in resultant ` `        ``// string 'result' by min(count1[\$i], ` `        ``// count2[\$i]) times ` `        ``for` `(``\$j` `= 1; ``\$j` `<= min(``\$count1``[``\$i``],  ` `                               ``\$count2``[``\$i``]); ``\$j``++) ` `            ``\$result` `= ``\$result``.``chr``(ord(``'a'``) + ``\$i``); ` ` `  `    ``echo` `\$result``; ` `} ` ` `  `// Driver Code ` `\$str1` `= ``"geeks"``; ` `\$str2` `= ``"cake"``; ` `longestString(``\$str1``, ``\$str2``); ` ` `  `// This code is contributed by ita_c ` `?> `

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.