# Count of subarrays having exactly K distinct elements

Given an array arr[] of size N and an integer K. The task is to find the count of subarrays such that each subarray has exactly K distinct elements.

Examples:

Input: arr[] = {2, 1, 2, 1, 6}, K = 2
Output: 7
{2, 1}, {1, 2}, {2, 1}, {1, 6}, {2, 1, 2},
{1, 2, 1} and {2, 1, 2, 1} are the only valid subarrays.

Input: arr[] = {1, 2, 3, 4, 5}, K = 1
Output: 5

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

Approach: To directly count the subarrays with exactly K different integers is hard but to find the count of subarrays with at most K different integers is easy. So the idea is to find the count of subarrays with at most K different integers, let it be C(K), and the count of subarrays with at most (K – 1) different integers, let it be C(K – 1) and finally take their difference, C(K) – C(K – 1) which is the required answer.
Count of subarrays with at most K different elements can be easily calculated through the sliding window technique. The idea is to keep expanding the right boundary of the window till the count of distinct elements in the window is less than or equal to K and when the count of distinct elements inside the window becomes more than K, start shrinking the window from the left till the count becomes less than or equal to K. Also for every expansion, keep counting the subarrays as right – left + 1 where right and left are the boundaries of the current window.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `#include ` `using` `namespace` `std; ` ` `  ` `  `// Function to return the count of subarrays ` `// with at most K distinct elements using ` `// the sliding window technique ` `int` `atMostK(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` ` `  `    ``// To store the result ` `    ``int` `count = 0; ` ` `  `    ``// Left boundary of window ` `    ``int` `left = 0; ` ` `  `    ``// Right boundary of window ` `    ``int` `right = 0; ` ` `  `    ``// Map to keep track of number of distinct ` `    ``// elements in the current window ` `    ``map<``int``,``int``> map; ` `    ``// Loop to calculate the count ` `    ``while` `(right < n) { ` ` `  `        ``// Calculating the frequency of each ` `        ``// element in the current window ` `        ``if` `(map.find(arr[right])==map.end()) ` `            ``map[arr[right]]=0; ` `        ``map[arr[right]]++; ` ` `  `        ``// Shrinking the window from left if the ` `        ``// count of distinct elements exceeds K ` `        ``while` `(map.size() > k) { ` `            ``map[arr[left]]= map[arr[left]] - 1; ` `            ``if` `(map[arr[left]] == 0) ` `                ``map.erase(arr[left]); ` `            ``left++; ` `        ``} ` ` `  `        ``// Adding the count of subarrays with at most ` `        ``// K distinct elements in the current window ` `        ``count += right - left + 1; ` `        ``right++; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Function to return the count of subarrays ` `// with exactly K distinct elements ` `int` `exactlyK(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` ` `  `    ``// Count of subarrays with exactly k distinct ` `    ``// elements is equal to the difference of the ` `    ``// count of subarrays with at most K distinct ` `    ``// elements and the count of subararys with ` `    ``// at most (K - 1) distinct elements ` `    ``return` `(atMostK(arr, n, k) - atMostK(arr, n, k - 1)); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 1, 2, 1, 6 }; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]); ` `    ``int` `k = 2; ` ` `  `    ``cout<<(exactlyK(arr, n, k)); ` `} ` ` `  `// This code is contributed by chitranayal `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `public` `class` `GfG { ` ` `  `    ``// Function to return the count of subarrays ` `    ``// with at most K distinct elements using ` `    ``// the sliding window technique ` `    ``private` `static` `int` `atMostK(``int` `arr[], ``int` `n, ``int` `k) ` `    ``{ ` ` `  `        ``// To store the result ` `        ``int` `count = ``0``; ` ` `  `        ``// Left boundary of window ` `        ``int` `left = ``0``; ` ` `  `        ``// Right boundary of window ` `        ``int` `right = ``0``; ` ` `  `        ``// Map to keep track of number of distinct ` `        ``// elements in the current window ` `        ``HashMap map = ``new` `HashMap<>(); ` ` `  `        ``// Loop to calculate the count ` `        ``while` `(right < n) { ` ` `  `            ``// Calculating the frequency of each ` `            ``// element in the current window ` `            ``map.put(arr[right], map.getOrDefault(arr[right], ``0``) + ``1``); ` ` `  `            ``// Shrinking the window from left if the ` `            ``// count of distinct elements exceeds K ` `            ``while` `(map.size() > k) { ` `                ``map.put(arr[left], map.get(arr[left]) - ``1``); ` `                ``if` `(map.get(arr[left]) == ``0``) ` `                    ``map.remove(arr[left]); ` `                ``left++; ` `            ``} ` ` `  `            ``// Adding the count of subarrays with at most ` `            ``// K distinct elements in the current window ` `            ``count += right - left + ``1``; ` `            ``right++; ` `        ``} ` `        ``return` `count; ` `    ``} ` ` `  `    ``// Function to return the count of subarrays ` `    ``// with exactly K distinct elements ` `    ``private` `static` `int` `exactlyK(``int` `arr[], ``int` `n, ``int` `k) ` `    ``{ ` ` `  `        ``// Count of subarrays with exactly k distinct ` `        ``// elements is equal to the difference of the ` `        ``// count of subarrays with at most K distinct ` `        ``// elements and the count of subararys with ` `        ``// at most (K - 1) distinct elements ` `        ``return` `(atMostK(arr, n, k) - atMostK(arr, n, k - ``1``)); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = { ``2``, ``1``, ``2``, ``1``, ``6` `}; ` `        ``int` `n = arr.length; ` `        ``int` `k = ``2``; ` ` `  `        ``System.out.print(exactlyK(arr, n, k)); ` `    ``} ` `} `

## Python3

 `# Python3 implementation of the above approach ` ` `  `# Function to return the count of subarrays  ` `# with at most K distinct elements using  ` `# the sliding window technique ` `def` `atMostK(arr, n, k) : ` `     `  `    ``# To store the result ` `    ``count ``=` `0` `     `  `    ``# Left boundary of window ` `    ``left ``=` `0` `     `  `    ``# Right boundary of window  ` `    ``right ``=` `0` `     `  `    ``# Map to keep track of number of distinct  ` `    ``# elements in the current window  ` `    ``map` `=` `{} ` ` `  `    ``# Loop to calculate the count  ` `    ``while``(right < n) : ` `         `  `        ``if` `arr[right] ``not` `in` `map` `: ` `            ``map``[arr[right]] ``=` `0` `         `  `        ``# Calculating the frequency of each  ` `        ``# element in the current window      ` `        ``map``[arr[right]] ``+``=` `1` `         `  `        ``# Shrinking the window from left if the  ` `        ``# count of distinct elements exceeds K  ` `        ``while``(``len``(``map``) > k) : ` `             `  `            ``if` `arr[left] ``not` `in` `map` `: ` `                ``map``[arr[left]] ``=` `0` `                 `  `            ``map``[arr[left]] ``-``=` `1` `             `  `            ``if` `map``[arr[left]] ``=``=` `0` `: ` `                ``del` `map``[arr[left]] ` `             `  `            ``left ``+``=` `1` `         `  `        ``# Adding the count of subarrays with at most  ` `        ``# K distinct elements in the current window  ` `        ``count ``+``=` `right ``-` `left ``+` `1` `        ``right ``+``=` `1` `     `  `    ``return` `count ` `     `  `# Function to return the count of subarrays  ` `# with exactly K distinct elements  ` `def` `exactlyK(arr, n, k) : ` `     `  `    ``# Count of subarrays with exactly k distinct  ` `    ``# elements is equal to the difference of the  ` `    ``# count of subarrays with at most K distinct  ` `    ``# elements and the count of subararys with  ` `    ``# at most (K - 1) distinct elements  ` `    ``return` `(atMostK(arr, n, k) ``-` `            ``atMostK(arr, n, k ``-` `1``)) ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"` `: ` `    ``arr ``=` `[``2``, ``1``, ``2``, ``1``, ``6``] ` `    ``n ``=` `len``(arr) ` `    ``k ``=` `2` `     `  `    ``print``(exactlyK(arr, n, k)) ` `     `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GfG  ` `{ ` ` `  `    ``// Function to return the count of subarrays ` `    ``// with at most K distinct elements using ` `    ``// the sliding window technique ` `    ``private` `static` `int` `atMostK(``int` `[]arr, ``int` `n, ``int` `k) ` `    ``{ ` ` `  `        ``// To store the result ` `        ``int` `count = 0; ` ` `  `        ``// Left boundary of window ` `        ``int` `left = 0; ` ` `  `        ``// Right boundary of window ` `        ``int` `right = 0; ` ` `  `        ``// Map to keep track of number of distinct ` `        ``// elements in the current window ` `        ``Dictionary<``int``, ``int``> map = ``new` `Dictionary<``int``, ``int``>(); ` ` `  `        ``// Loop to calculate the count ` `        ``while` `(right < n)  ` `        ``{ ` ` `  `            ``// Calculating the frequency of each ` `            ``// element in the current window ` `            ``if``(map.ContainsKey(arr[right])) ` `                ``map[arr[right]] = map[arr[right]] + 1;  ` `            ``else` `                ``map.Add(arr[right], 1); ` ` `  `            ``// Shrinking the window from left if the ` `            ``// count of distinct elements exceeds K ` `            ``while` `(map.Count > k) ` `            ``{ ` `                ``if``(map.ContainsKey(arr[left]))  ` `                ``{ ` `                    ``map[arr[left]] = map[arr[left]] - 1; ` `                    ``if` `(map[arr[left]] == 0) ` `                        ``map.Remove(arr[left]); ` `                ``} ` `                ``left++; ` `            ``} ` ` `  `            ``// Adding the count of subarrays with at most ` `            ``// K distinct elements in the current window ` `            ``count += right - left + 1; ` `            ``right++; ` `        ``} ` `        ``return` `count; ` `    ``} ` ` `  `    ``// Function to return the count of subarrays ` `    ``// with exactly K distinct elements ` `    ``private` `static` `int` `exactlyK(``int` `[]arr, ``int` `n, ``int` `k) ` `    ``{ ` ` `  `        ``// Count of subarrays with exactly k distinct ` `        ``// elements is equal to the difference of the ` `        ``// count of subarrays with at most K distinct ` `        ``// elements and the count of subararys with ` `        ``// at most (K - 1) distinct elements ` `        ``return` `(atMostK(arr, n, k) - atMostK(arr, n, k - 1)); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int` `[]arr = { 2, 1, 2, 1, 6 }; ` `        ``int` `n = arr.Length; ` `        ``int` `k = 2; ` ` `  `        ``Console.Write(exactlyK(arr, n, k)); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```7
```

Time Complexity: O(N)
Space Complexity: O(N)

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