使字符串平衡的最小附加子数

给定一个字符串str的小写字符，任务是找到需要添加到字符串中以使其平衡的最少字符数。当且仅当每个字符的出现次数相等时，才认为字符串是平衡的。

例子：

Input: str = “geeksforgeeks”
Output: 15
Add 2 ‘g’, 2 ‘k’, 2 ‘s’, 3 ‘f’, 3 ‘o’ and 3 ‘r’.

Input: str = “abcd”
Output: 0
The string is already balanced.

方法：为了最大程度地减少所需的添加量，必须使每个字符的频率等于最频繁出现的元素的频率。因此，首先，创建一个频率数组并找到给定字符串的所有字符的频率。现在，所需的答案将是每个字符的频率绝对差与频率数组中最大频率的绝对差之和。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
#define MAX 26
  
// Function to return the minimum additions
// required to balance the given string
int minimumAddition(string str, int len)
{
  
    // To store the frequency of
    // the characters of str
    int freq[MAX] = { 0 };
  
    // Update the frequency of the characters
    for (int i = 0; i < len; i++) {
        freq[str[i] - 'a']++;
    }
  
    // To store the maximum frequency from the array
    int maxFreq = *max_element(freq, freq + MAX);
  
    // To store the minimum additions required
    int minAddition = 0;
    for (int i = 0; i < MAX; i++) {
  
        // Every character's frequency must be
        // equal to the frequency of the most
        // frequently occurring character
        if (freq[i] > 0) {
            minAddition += abs(maxFreq - freq[i]);
        }
    }
  
    return minAddition;
}
  
// Driver code
int main()
{
    string str = "geeksforgeeks";
    int len = str.length();
  
    cout << minimumAddition(str, len);
  
    return 0;
}

Java

// Java implementation of the approach 
class GFG 
{
    final static int MAX = 26; 
      
    static int max_element(int freq[])
    {
        int max_ele = freq[0];
        for(int i = 0; i < MAX; i++)
        {
            if(max_ele < freq[i])
                max_ele = freq[i];
        }
        return max_ele;
    }
      
    // Function to return the minimum additions 
    // required to balance the given string 
    static int minimumAddition(String str, int len) 
    { 
      
        // To store the frequency of 
        // the characters of str 
        int freq[] = new int[MAX];
      
        // Update the frequency of the characters 
        for (int i = 0; i < len; i++) 
        { 
            freq[str.charAt(i) - 'a']++; 
        } 
      
        // To store the maximum frequency from the array 
        int maxFreq = max_element(freq); 
      
        // To store the minimum additions required 
        int minAddition = 0; 
        for (int i = 0; i < MAX; i++) 
        { 
      
            // Every character's frequency must be 
            // equal to the frequency of the most 
            // frequently occurring character 
            if (freq[i] > 0) 
            { 
                minAddition += Math.abs(maxFreq - freq[i]); 
            } 
        } 
        return minAddition; 
    } 
      
    // Driver code 
    public static void main (String[] args)
    { 
        String str = "geeksforgeeks"; 
        int len = str.length(); 
      
        System.out.println(minimumAddition(str, len)); 
    } 
}
  
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
MAX = 26
  
# Function to return the minimum additions
# required to balance the given str1ing
def minimumAddition(str1, Len):
  
    # To store the frequency of
    # the characters of str1
    freq = [0 for i in range(MAX)]
  
    # Update the frequency of the characters
    for i in range(Len):
        freq[ord(str1[i]) - ord('a')] += 1
  
    # To store the maximum frequency from the array
    maxFreq = max(freq)
  
    # To store the minimum additions required
    minAddition = 0
    for i in range(MAX):
  
        # Every character's frequency must be
        # equal to the frequency of the most
        # frequently occurring character
        if (freq[i] > 0):
            minAddition += abs(maxFreq - freq[i])
  
    return minAddition
  
# Driver code
str1 = "geeksforgeeks"
Len = len(str1)
  
print(minimumAddition(str1, Len))
  
# This code is contributed Mohit Kumar

C#

// C# implementation of the approach
using System;
      
class GFG 
{
    static int MAX = 26; 
      
    static int max_element(int []freq)
    {
        int max_ele = freq[0];
        for(int i = 0; i < MAX; i++)
        {
            if(max_ele < freq[i])
                max_ele = freq[i];
        }
        return max_ele;
    }
      
    // Function to return the minimum additions 
    // required to balance the given string 
    static int minimumAddition(String str, int len) 
    { 
      
        // To store the frequency of 
        // the characters of str 
        int []freq = new int[MAX];
      
        // Update the frequency of the characters 
        for (int i = 0; i < len; i++) 
        { 
            freq[str[i] - 'a']++; 
        } 
      
        // To store the maximum frequency from the array 
        int maxFreq = max_element(freq); 
      
        // To store the minimum additions required 
        int minAddition = 0; 
        for (int i = 0; i < MAX; i++) 
        { 
      
            // Every character's frequency must be 
            // equal to the frequency of the most 
            // frequently occurring character 
            if (freq[i] > 0) 
            { 
                minAddition += Math.Abs(maxFreq - freq[i]); 
            } 
        } 
        return minAddition; 
    } 
      
    // Driver code 
    public static void Main (String[] args)
    { 
        String str = "geeksforgeeks"; 
        int len = str.Length; 
      
        Console.WriteLine(minimumAddition(str, len)); 
    } 
}
  
// This code is contributed by 29AjayKumar

输出：

15