由相同字符组成的长度为 K 的子串的最大计数

给定一个字符串str和一个整数k 。任务是计算由相同字符组成的长度为k的子串的出现次数。可以有多个这样的长度为 k 的子串，选择出现次数最多的那个作为str的子串（不重叠）。
例子：

Input: str = “aaacaabbaa”, k = 2 
Output: 3 
“aa” and “bb” are the only sub-strings of length 2 that consist of the same characters. 
“bb” appears only once as a sub-string of str whereas “aa” appears thrice (which is the answer)
Input: str = “abab”, k = 2 
Output: 0 
 

方法：迭代所有从字符“a”到“Z”和计数长度为k的仅由当前字符的字符串显示为STR的子串的次数。最后打印这些计数的最大值。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the count
// of the required sub-strings
int maxSubStrings(string s, int k)
{
    int maxSubStr = 0, n = s.size();
 
    // Iterate over all characters
    for (int c = 0; c < 26; c++) {
        char ch = 'a' + c;
 
        // Count with current character
        int curr = 0;
        for (int i = 0; i <= n - k; i++) {
            if (s[i] != ch)
                continue;
            int cnt = 0;
            while (i < n && s[i] == ch && cnt != k) {
                i++;
                cnt++;
            }
            i--;
 
            // If the substring has a length k
            // then increment count with current character
            if (cnt == k)
                curr++;
        }
 
        // Update max count
        maxSubStr = max(maxSubStr, curr);
    }
    return maxSubStr;
}
 
// Driver Code
int main()
{
    string s = "aaacaabbaa";
    int k = 2;
    cout << maxSubStrings(s, k);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
     
// Function to return the count
// of the required sub-strings
static int maxSubStrings(String s, int k)
{
    int maxSubStr = 0, n = s.length();
 
    // Iterate over all characters
    for (int c = 0; c < 26; c++)
    {
        char ch = (char)((int)'a' + c);
 
        // Count with current character
        int curr = 0;
        for (int i = 0; i <= n - k; i++)
        {
            if (s.charAt(i) != ch)
                continue;
            int cnt = 0;
            while (i < n && s.charAt(i) == ch &&
                                        cnt != k)
            {
                i++;
                cnt++;
            }
            i--;
 
            // If the substring has a length
            //  k then increment count with
            // current character
            if (cnt == k)
                curr++;
        }
 
        // Update max count
        maxSubStr = Math.max(maxSubStr, curr);
    }
    return maxSubStr;
}
 
// Driver Code
public static void main(String []args)
{
    String s = "aaacaabbaa";
    int k = 2;
    System.out.println(maxSubStrings(s, k));
}
}
 
// This code is contributed by
// tufan_gupta2000

Python3

# Python 3 implementation of the approach
 
# Function to return the count
# of the required sub-strings
def maxSubStrings(s, k):
    maxSubStr = 0
    n = len(s)
 
    # Iterate over all characters
    for c in range(27):
        ch = chr(ord('a') + c)
 
        # Count with current character
        curr = 0
        for i in range(n - k):
            if (s[i] != ch):
                continue
            cnt = 0
            while (i < n and s[i] == ch and
                                   cnt != k):
                i += 1
                cnt += 1
     
            i -= 1
 
            # If the substring has a length k then
            # increment count with current character
            if (cnt == k):
                curr += 1
 
        # Update max count
        maxSubStr = max(maxSubStr, curr)
 
    return maxSubStr
 
# Driver Code
if __name__ == '__main__':
    s = "aaacaabbaa"
    k = 2
    print(maxSubStrings(s, k))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;
 
class GFG
{
         
    // Function to return the count
    // of the required sub-strings
    static int maxSubStrings(String s, int k)
    {
        int maxSubStr = 0, n = s.Length;
     
        // Iterate over all characters
        for (int c = 0; c < 26; c++)
        {
            char ch = (char)((int)'a' + c);
     
            // Count with current character
            int curr = 0;
            for (int i = 0; i <= n - k; i++)
            {
                if (s[i] != ch)
                    continue;
                int cnt = 0;
                while (i < n && s[i] == ch &&
                                cnt != k)
                {
                    i++;
                    cnt++;
                }
                i--;
     
                // If the substring has a length
                // k then increment count with
                // current character
                if (cnt == k)
                    curr++;
            }
     
            // Update max count
            maxSubStr = Math.Max(maxSubStr, curr);
        }
        return maxSubStr;
    }
     
    // Driver Code
    public static void Main()
    {
        string s = "aaacaabbaa";
        int k = 2;
        Console.WriteLine(maxSubStrings(s, k));
    }
}
 
// This code is contributed by Ryuga

Javascript

      // JavaScript implementation of the approach
      // Function to return the count
      // of the required sub-strings
      function maxSubStrings(s, k) {
        var maxSubStr = 0,
          n = s.length;
 
        // Iterate over all characters
        for (var c = 0; c < 26; c++) {
          var ch = String.fromCharCode("a".charCodeAt(0) + c);
 
          // Count with current character
          var curr = 0;
          for (var i = 0; i <= n - k; i++) {
            if (s[i] !== ch) continue;
            var cnt = 0;
            while (i < n && s[i] === ch && cnt !== k) {
              i++;
              cnt++;
            }
            i--;
 
            // If the substring has a length
            // k then increment count with
            // current character
            if (cnt === k) curr++;
          }
 
          // Update max count
          maxSubStr = Math.max(maxSubStr, curr);
        }
        return maxSubStr;
      }
 
      // Driver Code
      var s = "aaacaabbaa";
      var k = 2;
      document.write(maxSubStrings(s, k));
    

输出：

3

时间复杂度： O(n)，其中 n 是字符串的长度。

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