给定字符串中字符纠正的数量，使它们相等
给定三根弦 A 、 B 和 C 。每一个都是一串由小写英文字母组成的长度 N 。任务是通过执行一个操作使所有字符串相等，在这个操作中，给定字符串的任何字符都可以被任何其他字符替换，打印所需的最小操作数。
示例:
输入:A =“place”，B =“abcde”，C =“ply be” T3】输出:6 A =“place”，B =“abcde”，C =“ply be”。 我们可以通过执行以下六个操作，在最少的操作数内完成任务: 将 B 中的第一个字符改为‘p’。B 现在是“pbcde” 将 B 中的第二个字符改为‘l’。B 现在是“plcde” 将 B 和 C 中的第三个字符改为‘a’。b 和 C 现在分别是“plade”和“plabe”。 将 B 中的第四个字符改为‘c’。b 现在是“place” 将 C 中的第四个字符改为‘C’。C 现在是【地点】 输入: A =“游戏”，B =“游戏”，C =“游戏” 输出: 0
方法:运行一个循环，检查所有字符串的IthT5】字符是否相等，然后不需要任何操作。如果两个字符相等，则需要一个操作，如果三个字符都不同，则需要两个操作。
下面是上述方法的实现:
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 语言(一种计算机语言，尤用于创建网站)
// 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));
}
}
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?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.
?>
java 描述语言
<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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