Minimum Number of Manipulations required to make two Strings Anagram Without Deletion of Character - GeeksforGeeks

16 Nov, 2021
Given two strings s1 and s2, we need to find the minimum number of manipulations required to make two strings anagram without deleting any character. 
Note:- The anagram strings have same set of characters, sequence of characters can be different. If deletion of character is allowed and cost is given, refer to Minimum Cost To Make Two Strings IdenticalQuestion Source: Yatra.com Interview Experience | Set 7 
Examples: 
Input : 
       s1 = "aba"
       s2 = "baa"
Output : 0
Explanation: Both String contains identical characters

Input :
       s1 = "ddcf"
       s2 = "cedk"
Output : 2
Explanation : Here, we need to change two characters
in either of the strings to make them identical. We 
can change 'd' and 'f' in s1 or 'e' and 'k' in s2.
Assumption: Length of both the Strings is considered similar 
C++
Java
Python3
C#
PHP
Javascript
// C++ Program to find minimum number// of manipulations required to make// two strings identical#include <bits/stdc++.h>using namespace std;     // Counts the no of manipulations    // required    int countManipulations(string s1, string s2)    {                 int count = 0;         // store the count of character        int char_count[26];                 for (int i = 0; i < 26; i++)        {            char_count[i] = 0;        }         // iterate though the first String        // and update count        for (int i = 0; i < s1.length(); i++)            char_count[s1[i] - 'a']++;         // iterate through the second string        // update char_count.        // if character is not found in        // char_count then increase count        for (int i = 0; i < s2.length(); i++)        {            char_count[s2[i] - 'a']--;              }               for(int i = 0; i < 26; ++i)        {          if(char_count[i] != 0)          {            count+=abs(char_count[i]);          }        }        return count / 2;    }     // Driver code    int main()    {         string s1 = "ddcf";        string s2 = "cedk";                 cout<<countManipulations(s1, s2);    }  // This code is contributed by vt_m.
// Java Program to find minimum number of manipulations// required to make two strings identicalpublic class Similar_strings {     // Counts the no of manipulations required    static int countManipulations(String s1, String s2)    {        int count = 0;         // store the count of character        int char_count[] = new int[26];         // iterate though the first String and update        // count        for (int i = 0; i < s1.length(); i++)            char_count[s1.charAt(i) - 'a']++;                // iterate through the second string        // update char_count.        // if character is not found in char_count        // then increase count        for (int i = 0; i < s2.length(); i++)        {            char_count[s2.charAt(i) - 'a']--;        }               for(int i = 0; i < 26; ++i)        {          if(char_count[i] != 0)          {            count+= Math.abs(char_count[i]);          }        }                 return count / 2;    }     // Driver code    public static void main(String[] args)    {         String s1 = "ddcf";        String s2 = "cedk";        System.out.println(countManipulations(s1, s2));    }}
# Python3 Program to find minimum number# of manipulations required to make# two strings identical # Counts the no of manipulations# requireddef countManipulations(s1, s2):         count = 0     # store the count of character    char_count = [0] * 26         for i in range(26):        char_count[i] = 0     # iterate though the first String    # and update count    for i in range(len( s1)):        char_count[ord(s1[i]) -                   ord('a')] += 1     # iterate through the second string    # update char_count.    # if character is not found in    # char_count then increase count    for i in range(len(s2)):        char_count[ord(s2[i]) - ord('a')] -= 1             for i in range(26):        if char_count[i] != 0:            count += abs(char_count[i])              return count / 2 # Driver codeif __name__ == "__main__":     s1 = "ddcf"    s2 = "cedk"         print(countManipulations(s1, s2)) # This code is contributed by ita_c
// C# Program to find minimum number// of manipulations required to make// two strings identicalusing System; public class GFG {     // Counts the no of manipulations    // required    static int countManipulations(string s1,                                  string s2)    {        int count = 0;         // store the count of character        int []char_count = new int[26];         // iterate though the first String        // and update count        for (int i = 0; i < s1.Length; i++)            char_count[s1[i] - 'a']++;         // iterate through the second string        // update char_count.        // if character is not found in        // char_count then increase count        for (int i = 0; i < s2.Length; i++)            char_count[s2[i] - 'a']--;               for(int i = 0; i < 26; ++i)        {            if(char_count[i] != 0)            {              count+= Math.Abs(char_count[i]);            }        }                 return count / 2;    }     // Driver code    public static void Main()    {         string s1 = "ddcf";        string s2 = "cedk";                 Console.WriteLine(            countManipulations(s1, s2));    }} // This code is contributed by vt_m.
<?php// PHP Program to find minimum number// of manipulations required to make// two strings identical // Counts the no of manipulations// requiredfunction countManipulations($s1, $s2){    $count = 0;     // store the count of character    $char_count = array_fill(0, 26, 0);     // iterate though the first String    // and update count    for ($i = 0; $i < strlen($s1); $i++)        $char_count[ord($s1[$i]) -                    ord('a')] += 1;     // iterate through the second string    // update char_count.    // if character is not found in    // char_count then increase count    for ($i = 0; $i < strlen($s2); $i++)    {        $char_count[ord($s2[$i]) -                    ord('a')] -= 1;             }       for ($i = 0; $i < 26; $i++)    {      if($char_count[i]!=0)      {        $count+=abs($char_count[i]);      }    }    return ($count) / 2;} // Driver code$s1 = "ddcf";$s2 = "cedk"; echo countManipulations($s1, $s2); // This code is contributed by Ryuga?>
<script> // Javascript program to find minimum number// of manipulations required to make// two strings identical     // Counts the no of manipulations// requiredfunction countManipulations(s1, s2){    let count = 0;     // Store the count of character    let char_count = new Array(26);    for(let i = 0; i < char_count.length; i++)    {        char_count[i] = 0;    }         // Iterate though the first String and    // update count    for(let i = 0; i < s1.length; i++)        char_count[s1[i].charCodeAt(0) -                     'a'.charCodeAt(0)]++;           // Iterate through the second string    // update char_count.    // If character is not found in char_count    // then increase count    for(let i = 0; i < s2.length; i++)    {        char_count[s2[i].charCodeAt(0) -                     'a'.charCodeAt(0)]--;    }        for(let i = 0; i < 26; ++i)    {        if (char_count[i] != 0)        {            count += Math.abs(char_count[i]);        }    }    return count / 2;} // Driver codelet s1 = "ddcf";let s2 = "cedk"; document.write(countManipulations(s1, s2)); // This code is contributed by avanitrachhadiya2155     </script>
Output: 
2
Time Complexity: O(n), where n is the length of the string. 
This article is contributed by Sumit Ghosh and improved by Md Istakhar Ansari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@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.
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