计算所需的最少字符替换次数，以便给定字符串满足给定条件

给定一个由小写字母组成的长度为N的字符串S ，任务是计算需要替换以使给定字符串S特殊的最少字符数。

A string S is special if and only if for all possible pairs (S[i], S[j]) ( 1-based indexing ) where 1 ≤ i ≤ N/2 and N / 2 + 1 ≤ j ≤ N, one of the following condition must be true:

• S[i] < S[j]: For indices (i, j), each character in the left half (S[i]) is less than each character in the right half (S[j]).
• S[i] > S[j]: For index (i, j), each character in left half (S[i]) is greater than each character in the right half (S[j]).
• S[i] = S[j]: For index (i, j), each character in left half (S[i]) is equal to each character in the right half (S[j]).

例子：

Input: S = “aababc“ 
Output: 2
Explanation:
The left and right part of the string are Left= “aab”, Right=”abc” 
Operation 1: Change s[4] → d
Operation 2: Change s[5] → d
After applying the above changes, the resultant string is “aabddc” which satisfy all the conditions for special string.

Input: S = “aabaab” 
Output: 2
Explanation:
The left and right part of the string are Left=”aab” Right=”aab”
Operation 1: Change s[3] → a
Operation 2: Change s[6] → a
After applying the above changes, the resultant string is “aaaaaa” which satisfy all the conditions for special string.

方法：这个想法是观察到s[i]==s[j]只有 26 个字母的变化可以通过计算字符的频率来完成。请按照以下步骤解决上述问题：

• 存储左半部分和右半部分字符的频率分别存储在数组left[]和right[]中。
• 初始化变量计数以跟踪要进行的最小更改次数。
• 使用变量d遍历范围[0, 26]并根据以下情况更改字符：
  • 对于s[i]等于s[j]：使所有字符等于一个字符c，则 count 的值为(N – left – right) 。
  • 对于s[i]小于s[j]：
    • 最初，使用最大数量的更改，在左侧进行，让更改为N/2 。
    • 减去左侧它们≤d找到剩余的字符改变所有字符。
    • 添加在右侧其是等于d改变那些比d为变化更大的字符中的所有字符– =左和变化+ =右。
  • 对于s[i]大于s[j]：
    • 最初，使用最大数量的更改，在右侧进行，让更改为N/2 。
    • 减去右侧被≤d找到剩余的字符改变所有字符。
    • 在左侧添加等于 d 的所有字符以将这些字符更改为 > d。 (s[i]change += left和change -= right 。
• 上述情况下所有计数中的最小值是所需的结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include
using namespace std;
 
// Function that finds the minimum
// count of steps required to make
// the string special
int minChange(string s, int n)
{
     
    // Stores the frequency of the
    // left & right half of string
    int L[26] = {0};
    int R[26] = {0};
     
    // Find frequency of left half
    for(int i = 0; i < n / 2; i++)
    {
        char ch = s[i];
        L[ch - 'a']++;
    }
     
    // Find frequency of left half
    for(int i = n / 2; i < n; i++)
    {
        char ch = s[i];
        R[ch - 'a']++;
    }
     
    int count = n;
     
    // Make all characters equal
    // to character c
    for(char ch = 'a'; ch <= 'z'; ch++)
    {
        count = min(count,
                    n - L[ch - 'a'] -
                        R[ch - 'a']);
    }
     
    // Case 1: For s[i] < s[j]
    int change = n / 2;
    for(int d = 0; d + 1 < 26; d++)
    {
         
        // Subtract all the characters
        // on left side that are <=d
        change -= L[d];
         
        // Adding all characters on the
        // right side that same as d
        change += R[d];
         
        // Find minimum value of count
        count = min(count, change);
    }
     
    // Similarly for Case 2: s[i] > s[j]
    change = n / 2;
     
    for(int d = 0; d + 1 < 26; d++)
    {
        change -= R[d];
        change += L[d];
        count = min(change, count);
    }
     
    // Return the minimum changes
    return count;
}
 
// Driver Code
int main()
{
     
    // Given string S
    string S = "aababc";
 
    int N = S.length();
 
    // Function Call
    cout << minChange(S, N) << "\n";
}
 
// This code is contributed by sallagondaavinashreddy7

Java

// Java program for the above approach
 
import java.io.*;
import java.util.*;
class Main {
 
    // Function that finds the minimum
    // count of steps required to make
    // the string special
    public static int minChange(String s,
                                int n)
    {
 
        // Stores the frequency of the
        // left & right half of string
        int L[] = new int[26];
        int R[] = new int[26];
 
        // Find frequency of left half
        for (int i = 0; i < n / 2; i++) {
            char ch = s.charAt(i);
            L[ch - 'a']++;
        }
 
        // Find frequency of left half
        for (int i = n / 2; i < n; i++) {
            char ch = s.charAt(i);
            R[ch - 'a']++;
        }
        int count = n;
 
        // Make all characters equal
        // to character c
        for (char ch = 'a'; ch <= 'z'; ch++) {
 
            count = Math.min(
                count,
                n - L[ch - 'a']
                    - R[ch - 'a']);
        }
 
        // Case 1: For s[i] < s[j]
        int change = n / 2;
        for (int d = 0; d + 1 < 26; d++) {
 
            // Subtract all the characters
            // on left side that are <=d
            change -= L[d];
 
            // Adding all characters on the
            // right side that same as d
            change += R[d];
 
            // Find minimum value of count
            count = Math.min(count, change);
        }
 
        // Similarly for Case 2: s[i] > s[j]
        change = n / 2;
 
        for (int d = 0; d + 1 < 26; d++) {
            change -= R[d];
            change += L[d];
            count = Math.min(change, count);
        }
 
        // Return the minimum changes
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given string S
        String S = "aababc";
 
        int N = S.length();
 
        // Function Call
        System.out.println(minChange(S, N));
    }
}

Python3

# Python3 program for the
# above approach
 
# Function that finds the minimum
# count of steps required to make
# the string special
def minChange (s, n):
 
    # Stores the frequency of the
    # left & right half of string
    L = [0] * 26;
    R = [0] * 26;
 
    # Find frequency of left half
    for i in range(0, n // 2):
        ch = s[i];
        L[ord(ch) -
          ord('a')] += 1;
 
    # Find frequency of left half
    for i in range(n // 2, n):
        ch = s[i];
        R[ord(ch) -
          ord('a')] += 1;
 
    count = n;
 
    # Make all characters equal
    # to character c
    for ch in range(ord('a'),
                    ord('z')):
        count = min(count, n - L[ch - ord('a')] -
                               R[ch - ord('a')]);
 
    # Case 1: For s[i] < s[j]
    change = n / 2;
     
    for d in range(0, 25):
       
        # Subtract all the characters
        # on left side that are <=d
        change -= L[d];
 
        # Adding all characters on the
        # right side that same as d
        change += R[d];
 
        # Find minimum value of count
        count = min(count, change);
 
    # Similarly for Case 2: s[i] > s[j]
    change = n / 2;
 
    for d in range(0, 25):
        change -= R[d];
        change += L[d];
        count = min(change, count);
 
    # Return the minimum changes
    return int(count);
 
# Driver Code
if __name__ == '__main__':
   
    # Given string S
    S = "aababc";
 
    N = len(S);
 
    # Function Call
    print(minChange(S, N));
 
# This code is contributed by shikhasingrajput

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function that finds the minimum
// count of steps required to make
// the string special
public static int minChange(String s,
                            int n)
{
     
    // Stores the frequency of the
    // left & right half of string
    int []L = new int[26];
    int []R = new int[26];
 
    // Find frequency of left half
    for(int i = 0; i < n / 2; i++)
    {
        char ch = s[i];
        L[ch - 'a']++;
    }
 
    // Find frequency of left half
    for(int i = n / 2; i < n; i++)
    {
        char ch = s[i];
        R[ch - 'a']++;
    }
    int count = n;
 
    // Make all characters equal
    // to character c
    for(char ch = 'a'; ch <= 'z'; ch++)
    {
        count = Math.Min(count,
                         n - L[ch - 'a'] -
                             R[ch - 'a']);
    }
 
    // Case 1: For s[i] < s[j]
    int change = n / 2;
    for(int d = 0; d + 1 < 26; d++)
    {
         
        // Subtract all the characters
        // on left side that are <=d
        change -= L[d];
         
        // Adding all characters on the
        // right side that same as d
        change += R[d];
 
        // Find minimum value of count
        count = Math.Min(count, change);
    }
 
    // Similarly for Case 2: s[i] > s[j]
    change = n / 2;
 
    for(int d = 0; d + 1 < 26; d++)
    {
        change -= R[d];
        change += L[d];
        count = Math.Min(change, count);
    }
 
    // Return the minimum changes
    return count;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given string S
    String S = "aababc";
     
    int N = S.Length;
     
    // Function Call
    Console.WriteLine(minChange(S, N));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
      // JavaScript program for the above approach
       
      // Function that finds the minimum
      // count of steps required to make
      // the string special
      function minChange(s, n)
      {
        // Stores the frequency of the
        // left & right half of string
        var L = new Array(26).fill(0);
        var R = new Array(26).fill(0);
 
        // Find frequency of left half
        for (var i = 0; i < n / 2; i++) {
          var ch = s[i].charCodeAt(0);
          L[ch - "a".charCodeAt(0)]++;
        }
 
        // Find frequency of left half
        for (var i = n / 2; i < n; i++) {
          var ch = s[i].charCodeAt(0);
          R[ch - "a".charCodeAt(0)]++;
        }
        var count = n;
 
        // Make all characters equal
        // to character c
        for (var ch = "a".charCodeAt(0);
        ch <= "z".charCodeAt(0); ch++) {
          count = Math.min(
            count,
            n - L[ch - "a".charCodeAt(0)] -
            R[ch - "a".charCodeAt(0)]
          );
        }
 
        // Case 1: For s[i] < s[j]
        var change = parseInt(n / 2);
        for (var d = 0; d + 1 < 26; d++) {
          // Subtract all the characters
          // on left side that are <=d
          change -= L[d];
 
          // Adding all characters on the
          // right side that same as d
          change += R[d];
 
          // Find minimum value of count
          count = Math.min(count, change);
        }
 
        // Similarly for Case 2: s[i] > s[j]
        change = n / 2;
 
        for (var d = 0; d + 1 < 26; d++) {
          change -= R[d];
          change += L[d];
          count = Math.min(change, count);
        }
 
        // Return the minimum changes
        return count;
      }
 
      // Driver Code
      // Given string S
      var S = "aababc";
      var N = S.length;
 
      // Function Call
      document.write(minChange(S, N));
       

输出：

2

时间复杂度： O(N)
辅助空间： O(1)

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