使字符串K成为周期性的最小交换次数

给定长度为N的字符串S和由小写字母组成的数组A。还给定一个正整数K。任务是找到使字符串S K为周期性所需的最小交换次数(在S和A之间)。
笔记：

• 如果对于字符串S [i] = S [i + K]中的每个位置i，则该字符串都称为K周期。
• 在一个举动，S只有一个字符可以与A的字符互换
• A中的字符可以多次使用。

例子：

Input: S = “nihsiakyt”, K = 3, A = [‘n’, ‘i’, ‘p’, ‘s’, ‘q’] 
Output: 6 
Explanation: 
Considering 0 based positioning for the string S: 
Positions 3, 6 should be replaced with ‘n’. 
Position 7 should be replaced with ‘i’. 
Positions 2, 5, 8 can be replaced with any character from A to make the string K periodic. 
Final 3 periodic string: “nisnisnis” 
Therefore minimum swaps required = 6

Input: S = “abcdeactr”, K = 4, A = [‘a’, ‘c’, ‘p’] 
Output: 5 
Explanation: 
In total 5 changes are needed to make the string 4 periodic.

方法：可以通过频率计数和哈希处理解决此问题。

1. 为了解决上述问题，我们使用二维数组freq [K] [26]来存储所有位置j％K处的字符频率$0 \le j < N$ 。
2. 使用布尔数组标记数组A中存在的所有字符。
3. 对于0到K范围内的所有字符，将有N / K或(N / K + 1)个字符应该相同。
4. 因此，对于所有此类字符，我们检查哪个字符在位置i处具有最大频率，并且也出现在数组A中。
5. 将其添加到答案，即$(N / K - maxfrequency)$ 。
6. 如果满足以下条件，我们还将在答案中加1 $i % K < N % K$
7. 因为将会有N / K + 1字符的所有这样的字符我。

下面是上述方法的实现：

C++

// C++ code to find the minimum
// number of swaps required to
// make the string K periodic
  
#include 
using namespace std;
  
int minFlip(string s, int n,
            int k, char a[],
            int p)
{
    bool allowed[26] = { 0 };
  
    for (int i = 0; i < p; i++) {
  
        // Mark all allowed
        // characters as true
        allowed[a[i] - 'a'] = true;
    }
    char freq[k][26];
  
    // Initialize the freq array to 0
    for (int i = 0; i < k; i++)
        for (int j = 0; j < 26; j++)
            freq[i][j] = 0;
  
    for (int i = 0; i < n; i++) {
  
        // Increase the frequency
        // of each character
        freq[i % k][s[i] - 'a'] += 1;
    }
  
    int ans = 0;
  
    // Total number of periods of
    // size K in the string
    int totalpositions = n / k;
  
    for (int i = 0; i < k; i++) {
        int maxfrequency = 0;
  
        for (int j = 0; j < 26; j++) {
  
            // Check if the current character
            // is present in allowed
            // and whether the current
            // frequency is greater than
            // all previous frequencies
            // for this position
            if (freq[i][j] > maxfrequency
                and allowed[j] == true)
                maxfrequency = freq[i][j];
        }
  
        // update the answer by
        // subtracting the maxfrequency
        // from total positions
        // if there exist extra character
        // at the end of the string
        // apart from the n/k characters
        // then add 1.
        ans
            += (totalpositions
                - maxfrequency
                + ((i % k < n % k)
                       ? 1
                       : 0));
    }
    cout << ans << endl;
}
  
// Driver code
int main()
{
    string S = "nihsiakyt";
    int n = S.length();
  
    int K = 3;
  
    char A[5]
        = { 'n', 'i', 'p', 's', 'q' };
    int p = sizeof(A) / sizeof(A[0]);
  
    minFlip(S, n, K, A, p);
  
    return 0;
}

Java

// Java code to find the minimum
// number of swaps required to
// make the String K periodic
import java.util.*;
  
class GFG{
  
static void minFlip(String s, int n,
                    int k, char a[],
                    int p)
{
    boolean allowed[] = new boolean[26];
  
    for(int i = 0; i < p; i++)
    {
         
       // Mark all allowed
       // characters as true
       allowed[a[i] - 'a'] = true;
    }
    char [][]freq = new char[k][26];
  
    // Initialize the freq array to 0
    for(int i = 0; i < k; i++)
       for(int j = 0; j < 26; j++)
          freq[i][j] = 0;
  
    for(int i = 0; i < n; i++)
    {
         
       // Increase the frequency
       // of each character
       freq[i % k][s.charAt(i) - 'a'] += 1;
    }
  
    int ans = 0;
  
    // Total number of periods 
    // of size K in the String
    int totalpositions = n / k;
  
    for(int i = 0; i < k; i++)
    {
       int maxfrequency = 0;
       for(int j = 0; j < 26; j++)
       {
            
          // Check if the current character
          // is present in allowed
          // and whether the current
          // frequency is greater than
          // all previous frequencies
          // for this position
          if (freq[i][j] > maxfrequency && 
              allowed[j] == true)
              maxfrequency = freq[i][j];
       }
         
       // Update the answer by
       // subtracting the maxfrequency
       // from total positions
       // if there exist extra character
       // at the end of the String
       // apart from the n/k characters
       // then add 1.
       ans += (totalpositions - 
                 maxfrequency + 
                  ((i % k < n % 
                    k) ? 1 : 0));
    }
    System.out.print(ans + "\n");
}
  
// Driver code
public static void main(String[] args)
{
    String S = "nihsiakyt";
    int n = S.length();
    int K = 3;
  
    char []A = { 'n', 'i', 'p', 's', 'q' };
    int p = A.length;
  
    minFlip(S, n, K, A, p);
}
}
  
// This code is contributed by Amit Katiyar

Python3

# Python3 code to find the minimum
# number of swaps required to
# make the string K periodic
def minFlip(s, n, k, a, p):
  
    allowed = [0] * 26
  
    for i in range(p):
  
        # Mark all allowed
        # characters as true
        allowed[ord(a[i]) - ord('a')] = True
          
    freq = [[0 for x in range(26)]
               for y in range(k)]
  
    # Initialize the freq array to 0
    for i in range(k):
        for j in range(26):
            freq[i][j] = 0
  
    for i in range(n):
  
        # Increase the frequency
        # of each character
        freq[i % k][ord(s[i]) - ord('a')] += 1
  
    ans = 0
  
    # Total number of periods of
    # size K in the string
    totalpositions = n // k
  
    for i in range(k):
        maxfrequency = 0
        for j in range(26):
  
            # Check if the current character
            # is present in allowed
            # and whether the current
            # frequency is greater than
            # all previous frequencies
            # for this position
            if (freq[i][j] > maxfrequency and 
                allowed[j] == True):
                maxfrequency = freq[i][j]
  
        # Update the answer by
        # subtracting the maxfrequency
        # from total positions
        # if there exist extra character
        # at the end of the string
        # apart from the n/k characters
        # then add 1. 
        ans += (totalpositions - maxfrequency)
        if (i % k < n % k):
            ans += 1
          
    print(ans)
  
# Driver code
if __name__ == "__main__":
  
    S = "nihsiakyt"
    n = len(S)
  
    K = 3
  
    A = [ 'n', 'i', 'p', 's', 'q' ]
    p = len(A)
  
    minFlip(S, n, K, A, p)
  
# This code is contributed by chitranayal

C#

// C# code to find the minimum
// number of swaps required to
// make the String K periodic
using System;
  
class GFG{
  
static void minFlip(String s, int n,
                    int k, char []a,
                    int p)
{
    bool []allowed = new bool[26];
  
    for(int i = 0; i < p; i++)
    {
         
       // Mark all allowed
       // characters as true
       allowed[a[i] - 'a'] = true;
    }
    char [,]freq = new char[k, 26];
  
    // Initialize the freq array to 0
    for(int i = 0; i < k; i++)
       for(int j = 0; j < 26; j++)
          freq[i, j] = (char)0;
  
    for(int i = 0; i < n; i++)
    {
         
       // Increase the frequency
       // of each character
       freq[i % k, s[i] - 'a'] += (char)1;
    }
  
    int ans = 0;
  
    // Total number of periods 
    // of size K in the String
    int totalpositions = n / k;
  
    for(int i = 0; i < k; i++)
    {
       int maxfrequency = 0;
       for(int j = 0; j < 26; j++)
       {
  
          // Check if the current character
          // is present in allowed
          // and whether the current
          // frequency is greater than
          // all previous frequencies
          // for this position
          if (freq[i, j] > maxfrequency && 
              allowed[j] == true)
              maxfrequency = freq[i, j];
       }
         
       // Update the answer by
       // subtracting the maxfrequency
       // from total positions
       // if there exist extra character
       // at the end of the String
       // apart from the n/k characters
       // then add 1.
       ans += (totalpositions - 
                 maxfrequency + 
                  ((i % k < n % 
                    k) ? 1 : 0));
    }
    Console.Write(ans + "\n");
}
  
// Driver code
public static void Main(String[] args)
{
    String S = "nihsiakyt";
    int n = S.Length;
    int K = 3;
  
    char []A = { 'n', 'i', 'p', 's', 'q' };
    int p = A.Length;
  
    minFlip(S, n, K, A, p);
}
}
  
// This code is contributed by Rohit_ranjan

输出：

6