# Count of substrings of length K with exactly K distinct characters

Given a string str of lowercase alphabets and an integer K, the task is to count all substrings of length K which have exactly K distinct characters.

Example:

Input: str = “abcc”, K = 2
Output: 2
Explanation:
Possible substrings of length K = 2 are
ab : 2 distinct characters
bc : 2 distinct characters
cc : 1 distinct character
Only two valid substrings exist {“ab”, “bc”}.

Input: str = “aabab”, K = 3
Output: 0
Explanation:
Possible substrings of length K = 3 are
aab : 2 distinct characters
aba : 2 distinct characters
bab : 2 distinct characters
No substrings of length 3 exists with exactly 3 distinct characters

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

Naive approach:
The idea is to generate all substrings of length K and for each substring count number of distinct characters. If the length of a string is N, then there can be N – K + 1 substrings of length K. Generating these substrings will require O(N) complexity, and for checking each substring requires O(K) complexity, hence making the overall complexity as O(N*K).

Efficient approach:
The idea is to use Window Sliding Technique. Maintain a window of size K and keep a count of all the characters in the window using a HashMap. Traverse through the string reducing the count of the first character of the previous window and adding the frequency of the last character of the current window in the HashMap. If the count of distinct characters in a window of length K is equal to K, increment the answer by 1.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the ` `// count of k length substrings ` `// with k distinct characters ` `// using sliding window ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the ` `// required count of substrings ` `int` `countSubstrings(string str, ``int` `K) ` `{ ` `    ``int` `N = str.size(); ` `    ``// Store the count ` `    ``int` `answer = 0; ` ` `  `    ``// Store the count of ` `    ``// distinct characters ` `    ``// in every window ` `    ``unordered_map<``char``, ``int``> map; ` ` `  `    ``// Store the frequency of ` `    ``// the first K length substring ` `    ``for` `(``int` `i = 0; i < K; i++) { ` ` `  `        ``// Increase frequency of ` `        ``// i-th character ` `        ``map[str[i]]++; ` `    ``} ` ` `  `    ``// If K distinct characters ` `    ``// exist ` `    ``if` `(map.size() == K) ` `        ``answer++; ` ` `  `    ``// Traverse the rest of the ` `    ``// substring ` `    ``for` `(``int` `i = K; i < N; i++) { ` ` `  `        ``// Increase the frequency ` `        ``// of the last character ` `        ``// of the current substring ` `        ``map[str[i]]++; ` `        ``// Decrease the frequency ` `        ``// of the first character ` `        ``// of the previous substring ` `        ``map[str[i - K]]--; ` ` `  `        ``// If the character is not present ` `        ``// in the current substring ` `        ``if` `(map[str[i - K]] == 0) { ` `            ``map.erase(str[i - K]); ` `        ``} ` ` `  `        ``// If the count of distinct ` `        ``// characters is 0 ` `        ``if` `(map.size() == K) { ` `            ``answer++; ` `        ``} ` `    ``} ` ` `  `    ``// Return the count ` `    ``return` `answer; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// string str ` `    ``string str = ``"aabcdabbcdc"``; ` ` `  `    ``// integer K ` `    ``int` `K = 3; ` ` `  `    ``// Print the count of K length ` `    ``// substrings with k distinct characters ` `    ``cout << countSubstrings(str, K) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find the count ` `// of k length substrings with k  ` `// distinct characters using  ` `// sliding window  ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `// Function to return the  ` `// required count of substrings  ` `public` `static` `int` `countSubstrings(String str, ` `                                  ``int` `K)  ` `{  ` `    ``int` `N = str.length(); ` `     `  `    ``// Store the count  ` `    ``int` `answer = ``0``;  ` ` `  `    ``// Store the count of  ` `    ``// distinct characters  ` `    ``// in every window  ` `    ``Map map = ``new` `HashMap();  ` ` `  `    ``// Store the frequency of  ` `    ``// the first K length substring  ` `    ``for``(``int` `i = ``0``; i < K; i++)  ` `    ``{  ` `         `  `        ``// Increase frequency of  ` `        ``// i-th character  ` `        ``if` `(map.get(str.charAt(i)) == ``null``) ` `        ``{ ` `            ``map.put(str.charAt(i), ``1``); ` `        ``}  ` `        ``else` `        ``{ ` `            ``map.put(str.charAt(i),  ` `            ``map.get(str.charAt(i)) + ``1``); ` `        ``} ` `    ``}  ` ` `  `    ``// If K distinct characters  ` `    ``// exist  ` `    ``if` `(map.size() == K)  ` `        ``answer++;  ` ` `  `    ``// Traverse the rest of the  ` `    ``// substring  ` `    ``for``(``int` `i = K; i < N; i++) ` `    ``{  ` ` `  `        ``// Increase the frequency  ` `        ``// of the last character  ` `        ``// of the current substring  ` `        ``if` `(map.get(str.charAt(i)) == ``null``) ` `        ``{ ` `            ``map.put(str.charAt(i), ``1``); ` `        ``} ` `        ``else`  `        ``{ ` `            ``map.put(str.charAt(i),  ` `            ``map.get(str.charAt(i)) + ``1``); ` `        ``} ` `         `  `        ``// Decrease the frequency  ` `        ``// of the first character  ` `        ``// of the previous substring  ` `        ``map.put(str.charAt(i - K), ` `        ``map.get(str.charAt(i - K)) - ``1``); ` ` `  `        ``// If the character is not present  ` `        ``// in the current substring  ` `        ``if` `(map.get(str.charAt(i - K)) == ``0``) ` `        ``{  ` `            ``map.remove(str.charAt(i - K));  ` `        ``}  ` ` `  `        ``// If the count of distinct  ` `        ``// characters is 0  ` `        ``if` `(map.size() == K) ` `        ``{  ` `            ``answer++;  ` `        ``}  ` `    ``}  ` ` `  `    ``// Return the count  ` `    ``return` `answer;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `     `  `    ``// string str  ` `    ``String str = ``"aabcdabbcdc"``;  ` ` `  `    ``// integer K  ` `    ``int` `K = ``3``;  ` ` `  `    ``// Print the count of K length  ` `    ``// substrings with k distinct characters  ` `    ``System.out.println(countSubstrings(str, K)); ` `}  ` `} ` ` `  `// This code is contributed by grand_master `

## Python3

 `# Python3 program to find the ` `# count of k length substrings ` `# with k distinct characters ` `# using sliding window ` ` `  `# Function to return the ` `# required count of substrings ` `def` `countSubstrings(``str``, K): ` ` `  `    ``N ``=` `len``(``str``) ` ` `  `    ``# Store the count ` `    ``answer ``=` `0` ` `  `    ``# Store the count of ` `    ``# distinct characters ` `    ``# in every window ` `    ``map` `=` `{} ` ` `  `    ``# Store the frequency of ` `    ``# the first K length substring ` `    ``for` `i ``in` `range``(K): ` ` `  `        ``# Increase frequency of ` `        ``# i-th character ` `        ``map``[``str``[i]] ``=` `map``.get(``str``[i], ``0``) ``+` `1` `         `  `    ``# If K distinct characters ` `    ``# exist ` `    ``if` `(``len``(``map``) ``=``=` `K): ` `        ``answer ``+``=` `1` ` `  `    ``# Traverse the rest of the ` `    ``# substring ` `    ``for` `i ``in` `range``(K, N): ` ` `  `        ``# Increase the frequency ` `        ``# of the last character ` `        ``# of the current substring ` `        ``map``[``str``[i]] ``=` `map``.get(``str``[i], ``0``) ``+` `1` `         `  `        ``# Decrease the frequency ` `        ``# of the first character ` `        ``# of the previous substring ` `        ``map``[``str``[i ``-` `K]] ``-``=` `1` ` `  `        ``# If the character is not present ` `        ``# in the current substring ` `        ``if` `(``map``[``str``[i ``-` `K]] ``=``=` `0``): ` `            ``del` `map``[``str``[i ``-` `K]] ` ` `  `        ``# If the count of distinct ` `        ``# characters is 0 ` `        ``if` `(``len``(``map``) ``=``=` `K): ` `            ``answer ``+``=` `1` ` `  `    ``# Return the count ` `    ``return` `answer ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``str` `=` `"aabcdabbcdc"` ` `  `    ``# Integer K ` `    ``K ``=` `3` ` `  `    ``# Print the count of K length ` `    ``# substrings with k distinct characters ` `    ``print``(countSubstrings(``str``, K)) ` ` `  `# This code is contributed by mohit kumar 29 `

Output:

```5
```

Time Complexity: O(N) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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