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: 6
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 requiredconst 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 codeint 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 approachclass 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 requireddef 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 codeif __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 requiredfunction 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
Time Complexity: O(n)
Auxiliary Space: O(1)
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