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.

// 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.

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