# 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

## 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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