Maximum count of sub-strings of length K consisting of same characters

Given a string str and an integer k. The task is to count the occurrences of sub-strings of length k that consist of the same characters. There can be multiple such sub-strings possible of length k, choose the count of the one which appears the maximum number of times as the sub-string (non-overlapping) of str.

Examples: 

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 

Approach: Iterate over all the characters from 'a' to 'z' and count the number of times a string of length k consisting only of the current character appears as a sub-string of str. Print the maximum of these counts in the end.

Steps to solve the problem:

• Initialize maxSubStr to 0 and n to the size of the string s.
• Iterate over all characters of the English alphabet using a loop from 0 to 25.
• For each character, store it in a variable ch.
• Initialize a variable curr to 0.
• Use another loop to iterate over all possible substrings of length k in the string s.
• If the current character of the string s is not the same as the character ch, continue with the next iteration of the loop.
• If the current character is the same as ch, count the number of consecutive characters in the substring that are the same as ch, using a variable cnt.
• If cnt is not equal to k, continue with the next iteration of the loop.
• If cnt is equal to k, increment the variable curr.
• After the inner loop, update maxSubStr to the maximum of its current value and curr.
• After the outer loop, return maxSubStr as the result.

Below is the implementation of the above approach: 

// C++ implementation of the approach
#include <bits/stdc++.h>
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 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

# 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# 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

<script>
      // 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));
    </script>

3

Time Complexity: O(n), where n is the length of the string.
Auxiliary Space: O(1).