Longest common subsequence with permutations allowed - GeeksforGeeks

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"
 
The idea is to count characters in both strings.  
calculate frequency of characters for each string and store them in their respective count arrays, say count1[] for str1 and count2[] for str2.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.Since we traverse count array in ascending order, our final string characters will be in sorted order.
calculate frequency of characters for each string and store them in their respective count arrays, say count1[] for str1 and count2[] for str2.
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.
Since we traverse count array in ascending order, our final string characters will be in sorted order.
C++
Java
Python3
C#
PHP
Javascript
// 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 stringsvoid 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 caseint main(){    string str1 = "geeks", str2 = "cake";    longestString(str1, str2);    return 0;}
//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*/
# 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 stringsdef 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 Codeif __name__ == "__main__":         str1 = "geeks"    str2 = "cake"    longestString(str1, str2) # This code is contributed by ita_c
// C# program to find LCS with// permutations allowedusing 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 stringsstatic 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 Codepublic static void Main(){    String str1 = "geeks", str2 = "cake";    longestString(str1, str2);}} // This code is contributed// by PrinciRaj1992
<?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 stringsfunction 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?>
<script> // Javascript program to find LCS with permutations allowedfunction 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 stringsfunction 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>
ek
Time Complexity: O(m + n) where m and n are lengths of input strings.
Auxiliary Space: O(1)
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