最小化替换以使三个给定字符串中的任意两个相等

给定三个长度相等的字符串A 、 B和C ，任务是找到可以执行的最小替换操作数，以使三个字符串中的任何两个字符串相等。

例子：

Input: A = “aaa”, B = “bab”, C = “bbb” 
Output: 1 
Explanation: To make the strings B and C equal, B[1] = ‘a’ can be replaced with ‘b’. Hence the string B after replacement becomes B = “bbb” = C. Therefore, B and C can be made equal in 1 operation which is the minimum possible.

Input: A = “pqr”, B = “pqr”, C = “ppp” 
Output: 0 
Explanation: As A and B are already equal, no replacement operations are required. 

方法：给定问题可以通过找出所有三种可能情况的操作计数来解决，即A = B或B = C或A = C ，并取三者中的最小值。使字符串X和Y相等所需的操作数可以通过遍历字符串并跟踪索引使得X[i] ≠ Y[i]来计算。此类索引的计数是所需的操作数。

下面是上述方法的实现：

C++

// C++ program of the above approach
#include 
using namespace std;
 
// Function to find the minimum number of
// replacement to make two strings equal
int minimumReplacement(string X, string Y)
{
    // Stores the operation count
    int cnt = 0;
    for (int i = 0; i < X.length(); i++)
        if (X[i] != Y[i])
            cnt++;
 
    // Return Answer
    return cnt;
}
 
// Function to find the minimum number of
// replacement to make any two strings
// out of the given three strings equal
int replacementOperations(
    string A, string B, string C)
{
    // Case where A and B will be equal
    int AB = minimumReplacement(A, B);
 
    // Case where A and C will be equal
    int AC = minimumReplacement(A, C);
 
    // Case where B and C will be equal
    int BC = minimumReplacement(B, C);
 
    // Return the minimum of the three
    // above calculated values
    return min(AB, min(AC, BC));
}
 
// Driver Code
int main()
{
    string A = "aaa";
    string B = "bab";
    string C = "bbb";
 
    cout << replacementOperations(A, B, C);
 
    return 0;
}

Java

// Java program of the above approach
import java.util.*;
public class GFG
{
   
// Function to find the minimum number of
// replacement to make two strings equal
static int minimumReplacement(String X, String Y)
{
   
    // Stores the operation count
    int cnt = 0;
    for (int i = 0; i < X.length(); i++)
        if (X.charAt(i) != Y.charAt(i))
            cnt++;
 
    // Return Answer
    return cnt;
}
 
// Function to find the minimum number of
// replacement to make any two strings
// out of the given three strings equal
static int replacementOperations(
    String A, String B, String C)
{
    // Case where A and B will be equal
    int AB = minimumReplacement(A, B);
 
    // Case where A and C will be equal
    int AC = minimumReplacement(A, C);
 
    // Case where B and C will be equal
    int BC = minimumReplacement(B, C);
 
    // Return the minimum of the three
    // above calculated values
    return Math.min(AB, Math.min(AC, BC));
}
 
// Driver Code
public static void main(String args[])
{
    String A = "aaa";
    String B = "bab";
    String C = "bbb";
 
    System.out.println(replacementOperations(A, B, C));
}
}
 
// This code is contributed by Samim Hossain Mondal

Python3

# Python program of the above approach
 
# Function to find the minimum number of
# replacement to make two strings equal
def minimumReplacement(X, Y):
    # Stores the operation count
    cnt = 0
    for i in range(len(X)):
        if (X[i] != Y[i]):
            cnt += 1
 
    # Return Answer
    return cnt
 
# Function to find the minimum number of
# replacement to make any two strings
# out of the given three strings equal
def replacementOperations(A, B, C):
   
    # Case where A and B will be equal
    AB = minimumReplacement(A, B)
 
    # Case where A and C will be equal
    AC = minimumReplacement(A, C)
 
    # Case where B and C will be equal
    BC = minimumReplacement(B, C)
 
    # Return the minimum of the three
    # above calculated values
    return min(AB, min(AC, BC))
 
# Driver Code
A = "aaa"
B = "bab"
C = "bbb"
 
print(replacementOperations(A, B, C))
 
# This code is contributed by saurabh_jaiswal.

C#

// C# program of the above approach
using System;
using System.Collections;
 
class GFG
{
   
// Function to find the minimum number of
// replacement to make two strings equal
static int minimumReplacement(string X, string Y)
{
   
    // Stores the operation count
    int cnt = 0;
    for (int i = 0; i < X.Length; i++)
        if (X[i] != Y[i])
            cnt++;
 
    // Return Answer
    return cnt;
}
 
// Function to find the minimum number of
// replacement to make any two strings
// out of the given three strings equal
static int replacementOperations(
    string A, string B, string C)
{
   
    // Case where A and B will be equal
    int AB = minimumReplacement(A, B);
 
    // Case where A and C will be equal
    int AC = minimumReplacement(A, C);
 
    // Case where B and C will be equal
    int BC = minimumReplacement(B, C);
 
    // Return the minimum of the three
    // above calculated values
    return Math.Min(AB, Math.Min(AC, BC));
}
 
// Driver Code
public static void Main()
{
    string A = "aaa";
    string B = "bab";
    string C = "bbb";
 
    Console.Write(replacementOperations(A, B, C));
}
}
 
// This code is contributed by Samim Hossain Mondal

Javascript

// Javascript program of the above approach
 
 
// Function to find the minimum number of
// replacement to make two strings equal
function minimumReplacement(X, Y) {
  // Stores the operation count
  let cnt = 0;
  for (let i = 0; i < X.length; i++)
    if (X[i] != Y[i])
      cnt++;
 
  // Return Answer
  return cnt;
}
 
// Function to find the minimum number of
// replacement to make any two strings
// out of the given three strings equal
function replacementOperations(A, B, C) {
  // Case where A and B will be equal
  let AB = minimumReplacement(A, B);
 
  // Case where A and C will be equal
  let AC = minimumReplacement(A, C);
 
  // Case where B and C will be equal
  let BC = minimumReplacement(B, C);
 
  // Return the minimum of the three
  // above calculated values
  return Math.min(AB, Math.min(AC, BC));
}
 
// Driver Code
 
let A = "aaa";
let B = "bab";
let C = "bbb";
 
document.write(replacementOperations(A, B, C));

输出

1

时间复杂度： O(N)
辅助空间： O(1)