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Least number of manipulations needed to ensure two strings have identical characters

  • Difficulty Level : Easy
  • Last Updated : 30 Apr, 2021

Given two strings return the value of least number of manipulations needed to ensure both strings have identical characters, i.e., both string become anagram of each other.

Examples: 

Input : s1 = "aab"
        s2 = "aba" 
Output :  2
Explanation : string 1 contains 2 a's and 1 b, 
also string 2 contains same characters

Input : s1 = "abc"
        s2 = "cdd"
Output : 2
Explanation : string 1 contains 1 a, 1 b, 1 c
while string 2 contains 1 c and 2 d's
so there are 2 different characters

Question Source : Yatra.com Interview Experience | Set 7
 

The idea is to create a extra count array for both the strings separately and then count the difference in characters. 

C++




// C++ program to count least number
// of manipulations to have two strings
// set of same characters
#include <bits/stdc++.h>
using namespace std;
 
const int MAX_CHAR = 26;
 
// return the count of manipulations
// required
int leastCount(string s1, string s2, int n)
{
    int count1[MAX_CHAR] = { 0 };
    int count2[MAX_CHAR] = { 0 };
 
    // count the number of different
    // characters in both strings
    for (int i = 0; i < n; i++) {
        count1[s1[i] - 'a'] += 1;
        count2[s2[i] - 'a'] += 1;
    }
 
    // check the difference in characters
    // by comparing count arrays
    int res = 0;
    for (int i = 0; i < MAX_CHAR; i++) {
        if (count1[i] != 0) {
            res += abs(count1[i] - count2[i]);
        }
    }
    return res;
}
 
// driver program
int main()
{
    string s1 = "abc";
    string s2 = "cdd";
    int len = s1.length();
    int res = leastCount(s1, s2, len);
    cout << res << endl;
    return 0;
}

Java




// Java program to count least number
// of manipulations to have two
// strings set of same characters
import java.io.*;
 
public class GFG {
 
    static int MAX_CHAR = 26;
 
    // return the count of manipulations
    // required
    static int leastCount(String s1,
                        String s2, int n)
    {
         
        int[] count1 = new int[MAX_CHAR];
        int[] count2 = new int[MAX_CHAR];
 
        // count the number of different
        // characters in both strings
        for (int i = 0; i < n; i++)
        {
            count1[s1.charAt(i) - 'a'] += 1;
            count2[s2.charAt(i) - 'a'] += 1;
        }
 
        // check the difference in characters
        // by comparing count arrays
        int res = 0;
         
        for (int i = 0; i < MAX_CHAR; i++)
        {
            if (count1[i] != 0) {
                res += Math.abs(count1[i]
                                 - count2[i]);
            }
        }
         
        return res;
    }
 
    // driver program
    static public void main(String[] args)
    {
        String s1 = "abc";
        String s2 = "cdd";
        int len = s1.length();
        int res = leastCount(s1, s2, len);
         
        System.out.println(res);
    }
}
 
// This code is contributed by vt_m.

Python3




# Python3 program to count least number
# of manipulations to have two strings
# set of same characters
MAX_CHAR = 26
 
# return the count of manipulations
# required
def leastCount(s1, s2, n):
 
    count1 = [0] * MAX_CHAR
    count2 = [0] * MAX_CHAR
 
    # count the number of different
    # characters in both strings
    for i in range ( n):
        count1[ord(s1[i]) - ord('a')] += 1
        count2[ord(s2[i]) - ord('a')] += 1
 
    # check the difference in characters
    # by comparing count arrays
    res = 0
    for i in range (MAX_CHAR):
        if (count1[i] != 0):
            res += abs(count1[i] - count2[i])
     
    return res
 
# Driver Code
if __name__ == "__main__":
 
    s1 = "abc"
    s2 = "cdd"
    l = len(s1)
    res = leastCount(s1, s2, l)
    print (res)
 
# This code is contributed by ita_c

C#




// C# program to count least number
// of manipulations to have two strings
// set of same characters
using System;
 
public class GFG {
 
    static int MAX_CHAR = 26;
 
    // return the count of manipulations
    // required
    static int leastCount(string s1,
                        string s2, int n)
    {
         
        int[] count1 = new int[MAX_CHAR];
        int[] count2 = new int[MAX_CHAR];
 
        // count the number of different
        // characters in both strings
        for (int i = 0; i < n; i++)
        {
            count1[s1[i] - 'a'] += 1;
            count2[s2[i] - 'a'] += 1;
        }
 
        // check the difference in characters
        // by comparing count arrays
        int res = 0;
        for (int i = 0; i < MAX_CHAR; i++)
        {
            if (count1[i] != 0) {
                res += Math.Abs(count1[i]
                                - count2[i]);
            }
        }
         
        return res;
    }
 
    // driver program
    static public void Main()
    {
        string s1 = "abc";
        string s2 = "cdd";
        int len = s1.Length;
        int res = leastCount(s1, s2, len);
        Console.WriteLine(res);
    }
}
 
// This code is contributed by vt_m.

Javascript




<script>
 
// Javascript program to count least number
// of manipulations to have two
// strings set of same characters
let  MAX_CHAR = 26;
 
// Return the count of manipulations
// required
function leastCount(s1, s2, n)
{
    let count1 = new Array(MAX_CHAR);
    let count2 = new Array(MAX_CHAR);
     
    for(let i = 0; i < MAX_CHAR; i++)
    {
        count1[i] = 0;
        count2[i] = 0;
    }
     
    // Count the number of different
    // characters in both strings
    for(let i = 0; i < n; i++)
    {
        count1[s1[i].charCodeAt(0) -
                 'a'.charCodeAt(0)] += 1;
        count2[s2[i].charCodeAt(0) -
                 'a'.charCodeAt(0)] += 1;
    }
 
    // Check the difference in characters
    // by comparing count arrays
    let res = 0;
       
    for(let i = 0; i < MAX_CHAR; i++)
    {
        if (count1[i] != 0)
        {
            res += Math.abs(count1[i] - count2[i]);
        }
    }
    return res;
}
 
// Driver Code
let s1 = "abc";
let s2 = "cdd";
let len = s1.length;
let res = leastCount(s1, s2, len);
 
document.write(res);
 
// This code is contributed by avanitrachhadiya2155
     
</script>

Output: 

2

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