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

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. 

Implementation:

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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