要替换的最少字符以使所有字符的频率相同

给定一个由字母[‘A’ – ‘Z’]组成的字符串S ，任务是找到使每个字符的频率相等所需的最少操作次数。在一次操作中，可以选择字符串的任何字符并用另一个有效字符替换。
例子：

Input: S = “ABCB” 
Output: 1 
Explanation: 
In the given string character ‘C’ can be replaced by ‘A’, such that occurrence of every character becomes equal to 2. 
Updated String = “ABAB”
Input: S = “BBC” 
Output : 1 
Explanation: 
In the given string character ‘C’ can be replaced by ‘B’, such that occurrence of every character becomes equal to 3. 
Updated string = “BBB” 

方法：思路是找出字符串中每个字符出现的频率，然后根据出现的频率降序对字符进行排序。最后，我们可以检查字符串中产生最少要更改的字符数的每个字符，并打印要更改的最少字符数。
下面是上述方法的实现：

C++

// C++ implementation to find the
// Minimum characters to be replaced
// to make frequency of all characters same
 
#include 
using namespace std;
 
// Function to find the
// Minimum operations to convert
// given string to another with
// equal frequencies of characters
int minOperations(string s)
{
    // Frequency of characters
    int freq[26] = { 0 };
    int n = s.length();
 
    // Loop to find the Frequency
    // of each character
    for (int i = 0; i < n; i++) {
        freq[s[i] - 'A']++;
    }
 
    // Sort in decreasing order
    // based on frequency
    sort(freq, freq + 26, greater());
 
    // Maximum possible answer
    int answer = n;
 
    // Loop to find the minimum operations
    // required such that frequency of
    // every character is equal
    for (int i = 1; i <= 26; i++) {
        if (n % i == 0) {
            int x = n / i;
            int y = 0;
            for (int j = 0; j < i; j++) {
                y += min(freq[j], x);
            }
            answer = min(answer, n - y);
        }
    }
    return answer;
}
 
// Driver Code
int main()
{
    string s = "BBC";
    cout << minOperations(s);
 
    return 0;
}

Java

// Java implementation to find the
// Minimum characters to be replaced
// to make frequency of all characters same
  
 
import java.util.*;
 
class GFG{
  
// Function to find the
// Minimum operations to convert
// given String to another with
// equal frequencies of characters
static int minOperations(String s)
{
    // Frequency of characters
    Integer freq[] = new Integer[26];
    Arrays.fill(freq, 0);
    int n = s.length();
  
    // Loop to find the Frequency
    // of each character
    for (int i = 0; i < n; i++) {
        freq[s.charAt(i) - 'A']++;
    }
  
    // Sort in decreasing order
    // based on frequency
    Arrays.sort(freq, Collections.reverseOrder());
  
    // Maximum possible answer
    int answer = n;
  
    // Loop to find the minimum operations
    // required such that frequency of
    // every character is equal
    for (int i = 1; i <= 26; i++) {
        if (n % i == 0) {
            int x = n / i;
            int y = 0;
            for (int j = 0; j < i; j++) {
                y += Math.min(freq[j], x);
            }
            answer = Math.min(answer, n - y);
        }
    }
    return answer;
}
  
// Driver Code
public static void main(String[] args)
{
    String s = "BBC";
    System.out.print(minOperations(s));
  
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation to find the
# minimum characters to be replaced
# to make frequency of all characters same
 
# Function to find the minimum
# operations to convert given
# string to another with equal
# frequencies of characters
def minOperations(s):
 
    # Frequency of characters
    freq = [0] * 26
    n = len(s)
 
    # Loop to find the Frequency
    # of each character
    for i in range(n):
        freq[ord(s[i]) - ord('A')] += 1
     
    # Sort in decreasing order
    # based on frequency
    freq.sort(reverse = True)
 
    # Maximum possible answer
    answer = n
 
    # Loop to find the minimum operations
    # required such that frequency of
    # every character is equal
    for i in range(1, 27):
        if (n % i == 0):
            x = n // i
            y = 0
             
            for j in range(i):
                y += min(freq[j], x)
             
            answer = min(answer, n - y)
             
    return answer
 
# Driver Code
if __name__ == "__main__":
 
    s = "BBC"
     
    print (minOperations(s))
 
# This code is contributed by chitranayal

C#

// C# implementation to find the minimum
// characters to be replaced to make
// frequency of all characters same
using System;
 
class GFG{
 
// Function to find the minimum
// operations to convert given
// string to another with equal
// frequencies of characters
static int minOperations(String s)
{
     
    // Frequency of characters
    int []freq = new int[26];
    int n = s.Length;
 
    // Loop to find the frequency
    // of each character
    for(int i = 0; i < n; i++)
    {
       freq[s[i] - 'A']++;
    }
 
    // Sort in decreasing order
    // based on frequency
    Array.Sort(freq);
    Array.Reverse(freq);
     
    // Maximum possible answer
    int answer = n;
 
    // Loop to find the minimum operations
    // required such that frequency of
    // every character is equal
    for(int i = 1; i <= 26; i++)
    {
       if (n % i == 0)
       {
           int x = n / i;
           int y = 0;
            
           for(int j = 0; j < i; j++)
           {
              y += Math.Min(freq[j], x);
           }
           answer = Math.Min(answer, n - y);
       }
    }
    return answer;
}
 
// Driver Code
public static void Main(String[] args)
{
    String s = "BBC";
     
    Console.Write(minOperations(s));
 
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
      // JavaScript implementation to find the minimum
      // characters to be replaced to make
      // frequency of all characters same
       
      // Function to find the minimum
      // operations to convert given
      // string to another with equal
      // frequencies of characters
      function minOperations(s) {
        // Frequency of characters
        var freq = new Array(26).fill(0);
        var n = s.length;
 
        // Loop to find the frequency
        // of each character
        for (var i = 0; i < n; i++) {
          freq[s[i].charCodeAt(0) -
          "A".charCodeAt(0)]++;
        }
 
        // Sort in decreasing order
        // based on frequency
        freq.sort((a, b) => b - a);
 
        // Maximum possible answer
        var answer = n;
 
        // Loop to find the minimum operations
        // required such that frequency of
        // every character is equal
        for (var i = 1; i <= 26; i++) {
          if (n % i === 0) {
            var x = n / i;
            var y = 0;
 
            for (var j = 0; j < i; j++) {
              y += Math.min(freq[j], x);
            }
            answer = Math.min(answer, n - y);
          }
        }
        return answer;
      }
 
      // Driver Code
      var s = "BBC";
      document.write(minOperations(s));
       

输出：

1

时间复杂度： O(n)

辅助空间： O(26)

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