通过将元音更改为辅音来最小化更改以使所有字符相等，反之亦然

给定一个由小写字符组成的字符串str[] ，任务是使字符串的所有字符在最少的操作数中相等，以便在每个操作中选择一个元音并将其更改为辅音，反之亦然。

例子：

Input: str[] = “geeksforgeeks”
Output: 10
Explanation: To make all the characters equal, make the following changes – 

1. Change ‘o’ to a consonant(let’s say) ‘z’ and then to ‘e’
2. Change every other consonant(‘g’, ‘k’, ‘s’, ‘ f’, ‘r’, ) to ‘e’

This results in the string str = “eeeeeeeeeeeee” and the total number of operations performed is 10.

Input: str[] = “coding”
Output: 6

方法：从这个问题可以推导出辅音转元音需要1个操作，元音转元音或辅音转辅音需要2个步骤如果将元音变为元音或将辅音变为元音，然后再将辅音变为辅音，则将元音变为辅音，然后再返回元音。这个想法是保持两个变量 -

• Cv = 将所有字符更改为最大出现元音的成本 = 否。辅音数+（元音总数-最大元音出现的频率）* 2
• 抄送= 将所有字符更改为最大出现辅音的成本 = 否。元音+（辅音总数-最大出现辅音的频率）* 2

现在这两个中的最小值，即min(Cv, Cc)将给出我们可以转换字符串所需的最小步数。请按照以下步骤解决问题：

• 将变量ans、元音和辅音初始化为0以存储答案、元音个数和辅音个数。
• 将 2 个变量max_vowels和max_consonants初始化为INT_MIN以找到给定字符串中出现的最大元音和出现的最大辅音。
• 初始化 2 个 unordered_map freq_vowels[]和freq_consonant[]存储元音和辅音的频率。
• 使用变量i迭代范围[0, N)并执行以下步骤：
  • 如果当前字符是元音，则将元音的数量增加1 ，并将其在地图中的频率增加1 ，否则对辅音执行此操作。
• 遍历 unordered_maps 并找到出现的最大元音和辅音。
• 使用上面的公式，计算ans。
• 执行上述步骤后，打印ans的值作为答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the minimum number
// of steps to make all characters equal
int operations(string s)
{
 
    // Initializing the variables
    int ans = 0;
    int vowels = 0, consonants = 0;
    int max_vowels = INT_MIN;
    int max_consonants = INT_MIN;
 
    // Store the frequency
    unordered_map freq_consonants;
    unordered_map freq_vowels;
 
    // Iterate over the string
    for (int i = 0; i < s.size(); i++) {
        if (s[i] == 'a' or s[i] == 'e' or s[i] == 'i'
            or s[i] == 'o' or s[i] == 'u') {
 
            // Calculate the total
            // number of vowels
            vowels += 1;
 
            // Storing frequency of
            // each vowel
            freq_vowels[s[i]] += 1;
        }
        else {
 
            // Count the consonants
            consonants += 1;
 
            // Storing the frequency of
            // each consonant
            freq_consonants[s[i]] += 1;
        }
    }
 
    // Iterate over the 2 maps
    for (auto it = freq_consonants.begin();
         it != freq_consonants.end(); it++) {
 
        // Maximum frequency of consonant
        max_consonants = max(
            max_consonants,
            it->second);
    }
    for (auto it = freq_vowels.begin();
         it != freq_vowels.end(); it++) {
 
        // Maximum frequency of vowel
        max_vowels
            = max(max_vowels,
                  it->second);
    }
 
    // Find the result
    ans = min((2 * (vowels - max_vowels)
               + consonants),
              (2 * (consonants - max_vowels)
               + consonants));
 
    return ans;
}
 
// Driver Code
int main()
{
    string S = "geeksforgeeks";
    cout << operations(S);
 
    return 0;
}

Python3

# Python 3 program for the above approach
import sys
from collections import defaultdict
 
# Function to find the minimum number
# of steps to make all characters equal
def operations(s):
 
    # Initializing the variables
    ans = 0
    vowels = 0
    consonants = 0
    max_vowels = -sys.maxsize-1
    max_consonants = -sys.maxsize-1
 
    # Store the frequency
    freq_consonants = defaultdict(int)
    freq_vowels = defaultdict(int)
 
    # Iterate over the string
    for i in range(len(s)):
        if (s[i] == 'a' or s[i] == 'e' or s[i] == 'i'
                or s[i] == 'o' or s[i] == 'u'):
 
            # Calculate the total
            # number of vowels
            vowels += 1
 
            # Storing frequency of
            # each vowel
            freq_vowels[s[i]] += 1
 
        else:
 
            # Count the consonants
            consonants += 1
 
            # Storing the frequency of
            # each consonant
            freq_consonants[s[i]] += 1
 
    # Iterate over the 2 maps
    for it in freq_consonants:
 
        # Maximum frequency of consonant
        max_consonants = max(
            max_consonants,
            freq_consonants[it])
 
    for it in freq_vowels:
 
        # Maximum frequency of vowel
        max_vowels = max(max_vowels,
                         freq_vowels[it])
 
    # Find the result
    ans = min((2 * (vowels - max_vowels)
               + consonants),
              (2 * (consonants - max_vowels)
               + consonants))
 
    return ans
 
# Driver Code
if __name__ == "__main__":
 
    S = "geeksforgeeks"
    print(operations(S))
 
    # This code is contributed by ukasp.

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to find the minimum number
        // of steps to make all characters equal
        function operations(s) {
 
            // Initializing the variables
            let ans = 0;
            let vowels = 0, consonants = 0;
            let max_vowels = Number.MIN_VALUE;
            let max_consonants = Number.MIN_VALUE;
 
            // Store the frequency
            let freq_consonants = new Map();
            let freq_vowels = new Map();
 
            // Iterate over the string
            for (let i = 0; i < s.length; i++)
            {
                if (s.charAt(i) == 'a' || s.charAt(i) == 'e' || s.charAt(i) == 'i'
                    || s.charAt(i) == 'o' || s.charAt(i) == 'u') {
 
                    // Calculate the total
                    // number of vowels
                    vowels += 1;
 
                    // Storing frequency of
                    // each vowel
                    if (freq_vowels.has(s.charAt(i))) {
                        freq_vowels.set(s.charAt(i), freq_vowels.get(s.charAt(i)) + 1)
                    }
                    else {
                        freq_vowels.set(s.charAt(i), 1);
                    }
 
                }
                else {
 
                    // Count the consonants
                    consonants += 1;
 
                    // Storing the frequency of
                    // each consonant
                    if (freq_consonants.has(s.charAt(i))) {
                        freq_consonants.set(s.charAt(i), freq_consonants.get(s.charAt(i)) + 1)
                    }
                    else {
                        freq_consonants.set(s.charAt(i), 1);
                    }
 
                }
            }
 
            // Iterate over the 2 maps
            for (let [key, value] of freq_consonants) {
 
                // Maximum frequency of consonant
                max_consonants = Math.max(
                    max_consonants,
                    value);
            }
            for (let [key1, value1] of freq_vowels) {
 
                // Maximum frequency of vowel
                max_vowels
                    = Math.max(max_vowels,
                        value1);
            }
 
            // Find the result
            ans = Math.min((2 * (vowels - max_vowels)
                + consonants),
                (2 * (consonants - max_vowels)
                    + consonants));
 
            return ans;
        }
 
        // Driver Code
        let S = "geeksforgeeks";
        document.write(operations(S));
 
     // This code is contributed by Potta Lokesh
    

输出：

10

时间复杂度： O(N)
辅助空间： O(N)