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Maximum count of sub-strings of length K consisting of same characters
• Last Updated : 15 Feb, 2019

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

Below is the implementation of the above approach:

## 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`
Output:
```3
```

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

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