使字符串平衡的最小加法数

给定一个由小写字符组成的字符串str ，任务是找到需要添加到字符串中以使其平衡的最小字符数。当且仅当每个字符的出现次数相等时，才称字符串是平衡的。
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.

编程需要懂一点英语

方法：为了最小化所需的添加，每个字符的频率必须等于最频繁出现的元素的频率。所以首先，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，所需的答案将是频率数组中每个字符的频率与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 string
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

Javascript

输出：