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 any of the given strings can be replaced with any other character, print the count of 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 **i ^{th}** characters of all of 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 then 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 ` |

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## 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)); ` ` ` `} ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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. ` `?> ` |

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**Output:**

6

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