使字符串K 周期性所需的最小交换次数

给定一个长度为N的字符串S和一个由小写字母组成的数组A 。还给出了一个正整数K 。任务是找到使字符串S K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，则称该字符串是 K 周期的。
• 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：这个问题可以借助频率计数和哈希来解决。

1. 为了解决上面提到的问题，我们使用一个二维数组freq[K][26]来存储所有字符在j % K位置的频率$0 \le j < N$ .
2. 使用布尔数组来标记数组 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) 个应该相同的字符。
4. 所以对于所有这样的字符，我们检查哪个字符在位置 i 的频率最大，并且也出现在数组 A 中。
5. 将其添加到答案中，即$(N / K - maxfrequency)$ .
6. 我们还将在答案中加 1，如果$i % K < N % K$
7. 因为所有这样的字符i 都会有N / K + 1 个字符。

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

Javascript

 
// Javascript code for
// the above approach.
 
function minFlip(s, n, k, a, p)
{
    let allowed = Array.from({length: 26}, (_, i) => 0);
   
    for(let i = 0; i < p; i++)
    {
          
       // Mark all allowed
       // characters as true
       allowed[a[i].charCodeAt() - 'a'.charCodeAt()] = true;
    }
    let freq = new Array(k);
    for (var i = 0; i < freq.length; i++) {
    freq[i] = new Array(2);
    }
   
    // Initialize the freq array to 0
    for(let i = 0; i < k; i++)
       for(let j = 0; j < 26; j++)
          freq[i][j] = 0;
   
    for(let i = 0; i < n; i++)
    {
          
       // Increase the frequency
       // of each character
       freq[i % k][s[i].charCodeAt() - 'a'.charCodeAt()] += 1;
    }
   
    let ans = 0;
   
    // Total number of periods
    // of size K in the String
    let totalpositions = Math.floor(n / k);
   
    for(let i = 0; i < k; i++)
    {
       let maxfrequency = 0;
       for(let 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));
    }
    document.write(ans + "\n");
}
 
// Driver code
     
      let S = "nihsiakyt";
    let n = S.length;
    let K = 3;
   
    let A = [ 'n', 'i', 'p', 's', 'q' ];
    let p = A.length;
   
    minFlip(S, n, K, A, p);
                                                                                      

输出：

6

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。