Count of subarrays having exactly K distinct elements

20 Dec, 2021
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 
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++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include<bits/stdc++.h>#include<map>using namespace std;  // Function to return the count of subarrays// with at most K distinct elements using// the sliding window techniqueint 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    unordered_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 elementsint 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 codeint main(){    int arr[] = { 2, 1, 2, 1, 6 };    int n = sizeof(arr)/sizeof(arr[0]);    int k = 2;     cout<<(exactlyK(arr, n, k));}
// Java implementation of the approachimport 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<Integer, Integer> 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 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 codeif __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# implementation of the approachusing 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
<script> // Javascript implementation of the approach // Function to return the count of subarrays// with at most K distinct elements using// the sliding window techniquefunction atMostK(arr, n, k){    // To store the result    let count = 0;     // Left boundary of window    let left = 0;     // Right boundary of window    let right = 0;     // Map to keep track of number of distinct    // elements in the current window    let map = new Map();     // Loop to calculate the count    while (right < n)    {                 // Calculating the frequency of each        // element in the current window        if (map.has(arr[right]))            map.set(arr[right],            map.get(arr[right]) + 1);        else            map.set(arr[right], 1);         // Shrinking the window from left if the        // count of distinct elements exceeds K        while (map.size > k)        {            map.set(arr[left], map.get(arr[left]) - 1);            if (map.get(arr[left]) == 0)                map.delete(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 elementsfunction 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 codelet arr = [ 2, 1, 2, 1, 6 ];let n = arr.length;let k = 2; document.write(exactlyK(arr, n, k)); // This code is contributed by avanitrachhadiya2155 </script>
7
Time Complexity: O(N) Space Complexity: O(N)
Another Approach: When you move the right cursor, keep tracking whether we have reach a count of K distinct integers, if yes, we process left cursor, here is how we process left cursor:
check whether the element pointed by the left cursor is duplicated in the window, if yes, we remove it, and use a variable (e.g. prefix) to record that we have removed an element from the window). keep this process until we reduce the window size from to exactly K. now we can calculate the number of the valid good array as res += prefix;
after process left cursor and all the stuff, the outer loop will continue and the right cursor will move forward, and then the window size will exceed K, we can simply drop the leftmost element of the window and reset prefix to 0. and continue on.  
 
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ program to calculate number// of subarrays with distinct elements of size k#include <bits/stdc++.h>#include <map>#include <vector>using namespace std; int subarraysWithKDistinct(vector<int>& A, int K){       // declare a map for the frequency    unordered_map<int, int> mapp;    int begin = 0, end = 0, prefix = 0, cnt = 0;    int res = 0;       // traverse the array    while (end < A.size())    {        // increase the frequency        mapp[A[end]]++;        if (mapp[A[end]] == 1) {            cnt++;        }        end++;        if (cnt > K)        {            mapp[A[begin]]--;            begin++;            cnt--;            prefix = 0;        }               // loop until mapp[A[begin]] > 1        while (mapp[A[begin]] > 1)        {            mapp[A[begin]]--;            begin++;            prefix++;        }        if (cnt == K)        {            res += prefix + 1;        }    }       // return the final count    return res;}// Driver codeint main(){    vector<int> arr{ 2, 1, 2, 1, 6 };    int k = 2;      // Function call    cout << (subarraysWithKDistinct(arr, k));}// This code is contributed by Harman Singh
// Java program to calculate number// of subarrays with distinct elements of size kimport java.util.*;class GFG{  static int subarraysWithKDistinct(int A[], int K)  {     // declare a map for the frequency    HashMap<Integer, Integer> mapp = new HashMap<>();    int begin = 0, end = 0, prefix = 0, cnt = 0;    int res = 0;     // traverse the array    while (end < A.length)    {       // increase the frequency      if(mapp.containsKey(A[end]))      {        mapp.put(A[end], mapp.get(A[end]) + 1);      }      else      {        mapp.put(A[end], 1);      }      if (mapp.get(A[end]) == 1)      {        cnt++;      }      end++;      if (cnt > K)      {        if(mapp.containsKey(A[begin]))        {          mapp.put(A[begin], mapp.get(A[begin]) - 1);        }        else        {          mapp.put(A[begin], -1);        }        begin++;        cnt--;        prefix = 0;      }       // loop until mapp[A[begin]] > 1      while (mapp.get(A[begin]) > 1)      {        if(mapp.containsKey(A[begin]))        {          mapp.put(A[begin], mapp.get(A[begin]) - 1);        }        else        {          mapp.put(A[begin], -1);        }        begin++;        prefix++;      }      if (cnt == K)      {        res += prefix + 1;      }    }     // return the final count    return res;  }   // Driver code  public static void main(String[] args)  {    int arr[] = { 2, 1, 2, 1, 6 };    int k = 2;     // Function call    System.out.println(subarraysWithKDistinct(arr, k));  }} // This code is contributed by divyeshrabadiya07
# Python3 program to calculate number of# subarrays with distinct elements of size kdef subarraysWithKDistinct(A, K):         # Declare a map for the frequency    mapp = {}    begin, end, prefix, cnt = 0, 0, 0, 0    res = 0         # Traverse the array    while (end < len(A)):                 # Increase the frequency        mapp[A[end]] = mapp.get(A[end], 0) + 1                 if (mapp[A[end]] == 1):            cnt += 1         end += 1                 if (cnt > K):            mapp[A[begin]] -= 1            begin += 1            cnt -= 1            prefix = 0         # Loop until mapp[A[begin]] > 1        while (mapp[A[begin]] > 1):            mapp[A[begin]] -= 1            begin += 1            prefix += 1         if (cnt == K):            res += prefix + 1     # Return the final count    return res # Driver codeif __name__ == '__main__':         arr = [ 2, 1, 2, 1, 6 ]    k = 2         # Function call    print (subarraysWithKDistinct(arr, k))     # This code is contributed by Mohit kumar
// C# program to calculate number// of subarrays with distinct elements of size kusing System;using System.Collections.Generic;class GFG {         static int subarraysWithKDistinct(List<int> A, int K)    {                // declare a map for the frequency        Dictionary<int, int> mapp = new Dictionary<int, int>();         int begin = 0, end = 0, prefix = 0, cnt = 0;        int res = 0;                // traverse the array        while (end < A.Count)        {                       // increase the frequency            if(mapp.ContainsKey(A[end]))            {                mapp[A[end]]++;            }            else{                mapp[A[end]] = 1;            }            if (mapp[A[end]] == 1) {                cnt++;            }            end++;            if (cnt > K)            {                if(mapp.ContainsKey(A[begin]))                {                    mapp[A[begin]]--;                }                else{                    mapp[A[begin]] = -1;                }                begin++;                cnt--;                prefix = 0;            }                        // loop until mapp[A[begin]] > 1            while (mapp[A[begin]] > 1)            {                mapp[A[begin]]--;                begin++;                prefix++;            }            if (cnt == K)            {                res += prefix + 1;            }        }                // return the final count        return res;    }   // Driver code  static void Main()  {    List<int> arr = new List<int>(new int[] { 2, 1, 2, 1, 6 });    int k = 2;       // Function call    Console.Write(subarraysWithKDistinct(arr, k));  }} // This code is contributed by divyesh072019
<script> // Javascript program to calculate number// of subarrays with distinct elements of size kfunction subarraysWithKDistinct(A, K){         // Declare a map for the frequency    let mapp = new Map();    let begin = 0, end = 0, prefix = 0, cnt = 0;    let res = 0;      // Traverse the array    while (end < A.length)    {              // increase the frequency        if (mapp.has(A[end]))        {            mapp.set(A[end],            mapp.get(A[end]) + 1);        }        else        {            mapp.set(A[end], 1);        }        if (mapp.get(A[end]) == 1)        {            cnt++;        }        end++;        if (cnt > K)        {            if (mapp.has(A[begin]))            {                mapp.set(A[begin],                mapp.get(A[begin]) - 1);            }            else            {                mapp.set(A[begin], -1);            }            begin++;            cnt--;            prefix = 0;        }             // loop until mapp[A[begin]] > 1        while (mapp.get(A[begin]) > 1)        {            if(mapp.has(A[begin]))            {                mapp.set(A[begin],                mapp.get(A[begin]) - 1);            }            else            {                mapp.set(A[begin], -1);            }            begin++;            prefix++;        }        if (cnt == K)        {            res += prefix + 1;        }    }         // Return the final count    return res;} // Driver codelet arr = [ 2, 1, 2, 1, 6 ];let k = 2; // Function calldocument.write(subarraysWithKDistinct(arr, k)); // This code is contributed by rag2127 </script>
7
Time Complexity: O(N)Auxiliary Space: O(N)
29AjayKumar
ankthon
ukasp
harmansinghh
mohit kumar 29
divyesh072019
divyeshrabadiya07
avanitrachhadiya2155
rag2127
sooda367
pankajsharmagfg
ashutoshsinghgeeksforgeeks
amartyaghoshgfg
sliding-window
subarray
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Arrays
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