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Longest common subsequence with permutations allowed

  • Difficulty Level : Easy
  • Last Updated : 20 Jul, 2021

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"

 

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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.

C++




// 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;
}

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[2] --> 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*/

Python3




# 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[2] --> 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[2] --> 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];
    int []count2 = new int[26];
 
    // 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




<?php
// PHP 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[2] --> 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
?>

Javascript




<script>
 
// Javascript program to find LCS with permutations allowed
function min(a, b)
{
    if(a < b)
        return a;
       else
       return b;
}
 
// 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
function longestString( str1,  str2)
{
        var count1 = new Array(26);
        var count2 = new Array(26);
        count1.fill(0);
         count2.fill(0);
 
        // calculate frequency of characters
        for (var i = 0; i < str1.length; i++) {
            count1[str1.charCodeAt(i) -97]++;
        }
        for (var i = 0; i < str2.length; i++) {
            count2[str2.charCodeAt(i) - 97]++;
        }
 
        // Now traverse hash array
        var result = "";
        for (var i = 0; i < 26; i++)
         
        // append character ('a'+i) in resultant
        // String 'result' by min(count1[i],count2i])
        // times
        {
            for (var j = 1; j <= min(count1[i], count2[i]); j++) {
                result += String.fromCharCode(97 + i);
            }
        }
 
        document.write(result);
    }
        var str1 = "geeks";
        var str2 = "cake";
        longestString(str1, str2);
 
// This code is contributed by akshitsaxenaa09.
</script>
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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




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