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Number of character corrections in the given strings to make them equal

  • Difficulty Level : Medium
  • Last Updated : 11 May, 2021

Given three strings A, B, and C. Each of these is a string of length N consisting of lowercase English letters. The task is to make all the strings equal by performing an operation where any character of the given strings can be replaced with any other character, print the count of the minimum number of such operations required.

Examples:  

Input: A = “place”, B = “abcde”, C = “plybe” 
Output:
A = “place”, B = “abcde”, C = “plybe”. 
We can achieve the task in the minimum number of operations by performing six operations as follows: 
Change the first character in B to ‘p’. B is now “pbcde” 
Change the second character in B to ‘l’. B is now “plcde” 
Change the third character in B and C to ‘a’. B and C are now “plade” and “plabe” respectively. 
Change the fourth character in B to ‘c’. B is now “place” 
Change the fourth character in C to ‘c’. C is now “place”
Input: A = “game”, B = “game”, C = “game” 
Output: 0  

Approach: Run a loop, check if the ith characters of all the strings are equal then no operations are required. If two characters are equal then one operation is required and if all three characters are different from two operations are required. 
 

Below is the implementation of the above approach:  

C++




// C++ implementation of the approach
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of operations required
const int minOperations(int n, string a, string b, string c)
{
 
    // To store the count of operations
    int ans = 0;
    for (int i = 0; i < n; i++)
    {
        char x = a[i];
        char y = b[i];
        char z = c[i];
 
        // No operation required
        if (x == y && y == z)
            ;
 
        // One operation is required when
        // any two characters are equal
        else if (x == y || y == z || x == z)
        {
            ans++;
        }
             
        // Two operations are required when
        // none of the characters are equal
        else
        {
            ans += 2;
        }
    }
 
    // Return the minimum count of operations required
    return ans;
}
 
// Driver code
int main()
{
    string a = "place";
    string b = "abcde";
    string c = "plybe";
    int n = a.size();
    cout << minOperations(n, a, b, c);
    return 0;
}
 
// This code is contributed by 29AjayKumar

Java




// Java implementation of the approach
class GFG {
 
    // Function to return the count of operations required
    static int minOperations(int n, String a, String b, String c)
    {
 
        // To store the count of operations
        int ans = 0;
        for (int i = 0; i < n; i++) {
            char x = a.charAt(i);
            char y = b.charAt(i);
            char z = c.charAt(i);
 
            // No operation required
            if (x == y && y == z)
                ;
 
            // One operation is required when
            // any two characters are equal
            else if (x == y || y == z || x == z) {
                ans++;
            }
 
            // Two operations are required when
            // none of the characters are equal
            else {
                ans += 2;
            }
        }
 
        // Return the minimum count of operations required
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String a = "place";
        String b = "abcde";
        String c = "plybe";
        int n = a.length();
        System.out.print(minOperations(n, a, b, c));
    }
}

Python3




# Python 3 implementation of the approach
 
# Function to return the count
# of operations required
def minOperations(n, a, b, c):
     
    # To store the count of operations
    ans = 0
    for i in range(n):
        x = a[i]
        y = b[i]
        z = c[i]
 
        # No operation required
        if (x == y and y == z):
            continue
 
        # One operation is required when
        # any two characters are equal
        elif (x == y or y == z or x == z):
            ans += 1
             
        # Two operations are required when
        # none of the characters are equal
        else:
            ans += 2
 
    # Return the minimum count
    # of operations required
    return ans
 
# Driver code
if __name__ == '__main__':
    a = "place"
    b = "abcde"
    c = "plybe"
    n = len(a)
    print(minOperations(n, a, b, c))
 
# This code is contributed by
# Surendra_Gangwar

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the count of operations required
    static int minOperations(int n, string a, string b, string c)
    {
 
        // To store the count of operations
        int ans = 0;
        for (int i = 0; i < n; i++)
        {
            char x = a[i];
            char y = b[i];
            char z = c[i];
 
            // No operation required
            if (x == y && y == z)
                {;}
 
            // One operation is required when
            // any two characters are equal
            else if (x == y || y == z || x == z)
            {
                ans++;
            }
 
            // Two operations are required when
            // none of the characters are equal
            else
            {
                ans += 2;
            }
        }
 
        // Return the minimum count of operations required
        return ans;
    }
 
    // Driver code
    public static void Main()
    {
        string a = "place";
        string b = "abcde";
        string c = "plybe";
        int n = a.Length;
        Console.Write(minOperations(n, a, b, c));
    }
}
 
// This code is contributed by Ryuga

PHP




<?php
// PHP implementation of the approach
 
// Function to return the count of 
// operations required
function minOperations($n, $a, $b, $c)
{
 
    // To store the count of operations
    $ans = 0;
    for ($i = 0; $i < $n; $i++)
    {
        $x = $a[$i];
        $y = $b[$i];
        $z = $c[$i];
 
        // No operation required
        if ($x == $y && $y == $z)
            ;
 
        // One operation is required when
        // any two characters are equal
        else if ($x == $y ||
                 $y == $z || $x == $z)
        {
            $ans++;
        }
             
        // Two operations are required when
        // none of the characters are equal
        else
        {
            $ans += 2;
        }
    }
 
    // Return the minimum count of
    // operations required
    return $ans;
}
 
// Driver code
$a = "place";
$b = "abcde";
$c = "plybe";
$n = strlen($a);
echo minOperations($n, $a, $b, $c);
 
// This code is contributed by ajit.
?>

Javascript




<script>
    // Javascript implementation of the approach
     
    // Function to return the count of operations required
    function minOperations(n, a, b, c)
    {
   
        // To store the count of operations
        let ans = 0;
        for (let i = 0; i < n; i++)
        {
            let x = a[i];
            let y = b[i];
            let z = c[i];
   
            // No operation required
            if (x == y && y == z)
                {;}
   
            // One operation is required when
            // any two characters are equal
            else if (x == y || y == z || x == z)
            {
                ans++;
            }
   
            // Two operations are required when
            // none of the characters are equal
            else
            {
                ans += 2;
            }
        }
   
        // Return the minimum count of operations required
        return ans;
    }
     
    let a = "place";
    let b = "abcde";
    let c = "plybe";
    let n = a.length;
    document.write(minOperations(n, a, b, c));
 
</script>
Output: 
6

 

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