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

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 Identical

Question 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++

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// 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']--;
            if (char_count[s2[i] - 'a'] < 0) 
                count++; 
        }
        return count; 
    } 
  
    // Driver code 
    int main() 
    { 
  
        string s1 = "ddcf"; 
        string s2 = "cedk"; 
          
        cout<<countManipulations(s1, s2); 
    } 
   
// This code is contributed by vt_m.

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Java














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// Java Program to find minimum number of manipulations 


// required to make two strings identical 


public 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++)  


            if (char_count[s2.charAt(i) - 'a']-- <= 0) 


                count++; 


          


        return count; 


    } 


  


    // Driver code 


    public static void main(String[] args) 


    { 


  


        String s1 = "ddcf"; 


        String s2 = "cedk"; 


        System.out.println(countManipulations(s1, s2)); 


    } 


} 



















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Python 3














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# Python3 Program to find minimum number  


# of manipulations required to make  


# two strings identical  


  


# Counts the no of manipulations  


# required  


def 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


        if (char_count[ord(s2[i]) - 


                       ord('a')] < 0) : 


            count += 1


  


    return count 


  


# Driver code  


if __name__ == "__main__":  


  


    s1 = "ddcf"


    s2 = "cedk"


      


    print(countManipulations(s1, s2)) 


  


# This code is contributed by ita_c 



















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C#














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// C# Program to find minimum number 


// of manipulations required to make 


// two strings identical 


using 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++)  


            if (char_count[s2[i] - 'a']-- <= 0) 


                count++; 


          


        return count; 


    } 


  


    // Driver code 


    public static void Main() 


    { 


  


        string s1 = "ddcf"; 


        string s2 = "cedk"; 


          


        Console.WriteLine( 


            countManipulations(s1, s2)); 


    } 


} 


  


// This code is contributed by vt_m.  



















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PHP














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<?php 


// PHP Program to find minimum number  


// of manipulations required to make  


// two strings identical  


  


// Counts the no of manipulations  


// required  


function 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;  


          


        if ($char_count[ord($s2[$i]) -  


                        ord('a')] < 0)  


            $count++;  


    }  


    return $count;  


}  


  


// Driver code 


$s1 = "ddcf";  


$s2 = "cedk";  


  


echo countManipulations($s1, $s2);  


  


// This code is contributed by Ryuga 


?> 



















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Output:


2



Time Complexity : O(n), where n is the length of the string.

This article is contributed by Sumit Ghosh. 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.


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