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Number of substrings with count of each character as k
• Difficulty Level : Easy
• Last Updated : 01 May, 2021

Given a string and an integer k, find number of substrings in which all the different characters occurs exactly k times.

Examples:

```Input : s = "aabbcc"
k = 2
Output : 6
The substrings are aa, bb, cc,
aabb, bbcc and aabbcc.

Input : s = "aabccc"
k = 2
Output : 3
There are three substrings aa,
cc and cc```

The idea is to traverse through all substrings. We fix a starting point, traverse through all substrings starring with the picked point, we keep incrementing frequencies of all characters. If all frequencies become k, we increment result. If count of any frequency becomes more than k, we break and change starting point.

## C++

 `// C++ program to count number of substrings``// with counts of distinct characters as k.``#include ``using` `namespace` `std;``const` `int` `MAX_CHAR = 26;` `// Returns true if all values``// in freq[] are either 0 or k.``bool` `check(``int` `freq[], ``int` `k)``{``    ``for` `(``int` `i = 0; i < MAX_CHAR; i++)``        ``if` `(freq[i] && freq[i] != k)``            ``return` `false``;``    ``return` `true``;``}` `// Returns count of substrings where frequency``// of every present character is k``int` `substrings(string s, ``int` `k)``{``    ``int` `res = 0;  ``// Initialize result` `    ``// Pick a starting point``    ``for` `(``int` `i = 0; s[i]; i++) {` `        ``// Initialize all frequencies as 0``        ``// for this starting point``        ``int` `freq[MAX_CHAR] = { 0 };` `        ``// One by one pick ending points``        ``for` `(``int` `j = i; s[j]; j++) {`` ` `            ``// Increment frequency of current char``            ``int` `index = s[j] - ``'a'``;``            ``freq[index]++;` `            ``// If frequency becomes more than``            ``// k, we can't have more substrings``            ``// starting with i``            ``if` `(freq[index] > k)``                ``break``;` `            ``// If frequency becomes k, then check``            ``// other frequencies as well.``            ``else` `if` `(freq[index] == k &&``                  ``check(freq, k) == ``true``)``                ``res++;``        ``}``    ``}``    ``return` `res;``}` `// Driver code``int` `main()``{``    ``string s = ``"aabbcc"``;``    ``int` `k = 2;``    ``cout << substrings(s, k) << endl;` `    ``s = ``"aabbc"``;``    ``k = 2;``    ``cout << substrings(s, k) << endl;``}`

## Java

 `// Java program to count number of substrings``// with counts of distinct characters as k.``class` `GFG``{` `static` `int` `MAX_CHAR = ``26``;` `// Returns true if all values``// in freq[] are either 0 or k.``static` `boolean` `check(``int` `freq[], ``int` `k)``{``    ``for` `(``int` `i = ``0``; i < MAX_CHAR; i++)``        ``if` `(freq[i] !=``0` `&& freq[i] != k)``            ``return` `false``;``    ``return` `true``;``}` `// Returns count of substrings where frequency``// of every present character is k``static` `int` `substrings(String s, ``int` `k)``{``    ``int` `res = ``0``; ``// Initialize result` `    ``// Pick a starting point``    ``for` `(``int` `i = ``0``; i< s.length(); i++)``    ``{` `        ``// Initialize all frequencies as 0``        ``// for this starting point``        ``int` `freq[] = ``new` `int``[MAX_CHAR];` `        ``// One by one pick ending points``        ``for` `(``int` `j = i; j k)``                ``break``;` `            ``// If frequency becomes k, then check``            ``// other frequencies as well.``            ``else` `if` `(freq[index] == k &&``                ``check(freq, k) == ``true``)``                ``res++;``        ``}``    ``}``    ``return` `res;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``String s = ``"aabbcc"``;``    ``int` `k = ``2``;``    ``System.out.println(substrings(s, k));` `    ``s = ``"aabbc"``;``    ``k = ``2``;``    ``System.out.println(substrings(s, k));``}``}` `// This code has been contributed by 29AjayKumar`

## Python3

 `# Python3 program to count number of substrings``# with counts of distinct characters as k.` `MAX_CHAR ``=` `26` `# Returns true if all values``# in freq[] are either 0 or k.``def` `check(freq, k):``    ``for` `i ``in` `range``(``0``, MAX_CHAR):``        ``if``(freq[i] ``and` `freq[i] !``=` `k):``            ``return` `False``    ``return` `True` `# Returns count of substrings where``# frequency of every present character is k``def` `substrings(s, k):``    ``res ``=` `0` `# Initialize result` `    ``# Pick a starting point``    ``for` `i ``in` `range``(``0``, ``len``(s)):` `        ``# Initialize all frequencies as 0``        ``# for this starting point``        ``freq ``=` `[``0``] ``*` `MAX_CHAR` `        ``# One by one pick ending points``        ``for` `j ``in` `range``(i, ``len``(s)):``            ` `            ``# Increment frequency of current char``            ``index ``=` `ord``(s[j]) ``-` `ord``(``'a'``)``            ``freq[index] ``+``=` `1` `            ``# If frequency becomes more than``            ``# k, we can't have more substrings``            ``# starting with i``            ``if``(freq[index] > k):``                ``break``            ` `            ``# If frequency becomes k, then check``            ``# other frequencies as well``            ``elif``(freq[index] ``=``=` `k ``and``                 ``check(freq, k) ``=``=` `True``):``                ``res ``+``=` `1``            ` `    ``return` `res` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``s ``=` `"aabbcc"``    ``k ``=` `2``    ``print``(substrings(s, k))` `    ``s ``=` `"aabbc"``;``    ``k ``=` `2``;``    ``print``(substrings(s, k))` `# This code is contributed``# by Sairahul Jella`

## C#

 `// C# program to count number of substrings``// with counts of distinct characters as k.``using` `System;` `class` `GFG``{` `static` `int` `MAX_CHAR = 26;` `// Returns true if all values``// in freq[] are either 0 or k.``static` `bool` `check(``int` `[]freq, ``int` `k)``{``    ``for` `(``int` `i = 0; i < MAX_CHAR; i++)``        ``if` `(freq[i] != 0 && freq[i] != k)``            ``return` `false``;``    ``return` `true``;``}` `// Returns count of substrings where frequency``// of every present character is k``static` `int` `substrings(String s, ``int` `k)``{``    ``int` `res = 0; ``// Initialize result` `    ``// Pick a starting point``    ``for` `(``int` `i = 0; i < s.Length; i++)``    ``{` `        ``// Initialize all frequencies as 0``        ``// for this starting point``        ``int` `[]freq = ``new` `int``[MAX_CHAR];` `        ``// One by one pick ending points``        ``for` `(``int` `j = i; j < s.Length; j++)``        ``{` `            ``// Increment frequency of current char``            ``int` `index = s[j] - ``'a'``;``            ``freq[index]++;` `            ``// If frequency becomes more than``            ``// k, we can't have more substrings``            ``// starting with i``            ``if` `(freq[index] > k)``                ``break``;` `            ``// If frequency becomes k, then check``            ``// other frequencies as well.``            ``else` `if` `(freq[index] == k &&``                ``check(freq, k) == ``true``)``                ``res++;``        ``}``    ``}``    ``return` `res;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``String s = ``"aabbcc"``;``    ``int` `k = 2;``    ``Console.WriteLine(substrings(s, k));` `    ``s = ``"aabbc"``;``    ``k = 2;``    ``Console.WriteLine(substrings(s, k));``}``}` `/* This code contributed by PrinciRaj1992 */`

## PHP

 ` ``\$k``)``                ``break``;` `            ``// If frequency becomes k, then check``            ``// other frequencies as well.``            ``else` `if` `(``\$freq``[``\$index``] == ``\$k` `&&``                ``check(``\$freq``, ``\$k``) == true)``                ``\$res``++;``        ``}``    ``}``    ``return` `\$res``;``}` `// Driver code``\$s` `= ``"aabbcc"``;``\$k` `= 2;``echo` `substrings(``\$s``, ``\$k``).``"\n"``;``\$s` `= ``"aabbc"``;``\$k` `= 2;``echo` `substrings(``\$s``, ``\$k``).``"\n"``;` `// This code is contributed by Ita_c.``?>`

## Javascript

 ``

Output:

```6
3```

Time Complexity : O(n3) where n is length of input string.

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