Count distinct elements in every window of size k

Given an array of size n and an integer k, return the of count of distinct numbers in all windows of size k.

Example:

Input:  arr[] = {1, 2, 1, 3, 4, 2, 3};
            k = 4
Output:
3
4
4
3

Explanation:
First window is {1, 2, 1, 3}, count of distinct numbers is 3
Second window is {2, 1, 3, 4} count of distinct numbers is 4
Third window is {1, 3, 4, 2} count of distinct numbers is 4
Fourth window is {3, 4, 2, 3} count of distinct numbers is 3

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Source: Microsoft Interview Question

A Simple Solution is to traverse the given array, consider every window in it and count distinct elements in the window. Below is C++ implementation of simple solution.

// Simple C++ program to count distinct elements in every // window of size k #include <iostream> using namespace std;   // Counts distinct elements in window of size k int countWindowDistinct(int win[], int k) {     int dist_count = 0;       // Traverse the     for (int i=0; i<k; i++)     {         // Check if element arr[i] exists in arr[0..i-1]         int j;         for (j=0; j<i; j++)            if (win[i] == win[j])               break;         if (j==i)             dist_count++;     }     return dist_count; }   // Counts distinct elements in all windows of size k void countDistinct(int arr[], int n, int k) {     // Traverse through every window     for (int i=0; i<=n-k; i++)        cout << countWindowDistinct(arr+i, k) << endl; }   // Driver program int main() {     int arr[] = {1, 2, 1, 3, 4, 2, 3},  k = 4;     int n = sizeof(arr)/sizeof(arr[0]);     countDistinct(arr, n, k);     return 0; }

Output:

3
4
4
3

Time complexity of the above solution is O(nk2). We can improve time complexity to O(nkLok) by modifying countWindowDistinct() to use sorting. The function can further be optimized to use hashing to find distinct elements in a window. With hashing the time complexity becomes O(nk). Below is a different approach that works in O(n) time.

An Efficient Solution is to use the count of previous window, while sliding the window. The idea is to create a hash map that stores elements of current widow. When we slide the window, we remove an element from hash and add an element. We also keep track of distinct elements. Below is algorithm.

1) Create an empty hash map. Let hash map be hM

2) Initialize distinct element count ‘dist_count’ as 0.

3) Traverse through first window and insert elements of first window to hM. The elements are used as key and their counts as value in hM. Also keep updating ‘dist_count’

4) Print ‘dist_count’ for first window.

3) Traverse through remaining array (or other windows).
….a) Remove the first element of previous window.
…….If the removed element appeared only once
…………..remove it from hM and do “dist_count–”
…….Else (appeared multiple times in hM)
…………..decrement its count in hM

….a) Add the current element (last element of new window)
…….If the added element is not present in hM
…………..add it to hM and do “dist_count++”
…….Else (the added element appeared multiple times)
…………..increment its count in hM

Below is Java implementation of above approach.

// An efficient Java program to count distinct elements in // every window of size k import java.util.HashMap;   class CountDistinctWindow {     static void countDistinct(int arr[], int k)     {         // Creates an empty hashMap hM         HashMap<Integer, Integer> hM =                       new HashMap<Integer, Integer>();           // initialize distinct element  count for         // current window         int dist_count = 0;           // Traverse the first window and store count         // of every element in hash map         for (int i = 0; i < k; i++)         {             if (hM.get(arr[i]) == null)             {                 hM.put(arr[i], 1);                 dist_count++;             }             else             {                int count = hM.get(arr[i]);                hM.put(arr[i], count+1);             }         }           // Print count of first window         System.out.println(dist_count);           // Traverse through the remaining array         for (int i = k; i < arr.length; i++)         {               // Remove first element of previous window             // If there was only one occurrence, then             // reduce distinct count.             if (hM.get(arr[i-k]) == 1)             {                 hM.remove(arr[i-k]);                 dist_count--;             }             else // reduce count of the removed element             {                int count = hM.get(arr[i-k]);                hM.put(arr[i-k], count-1);             }               // Add new element of current window             // If this element appears first time,             // increment distinct element count             if (hM.get(arr[i]) == null)             {                 hM.put(arr[i], 1);                 dist_count++;             }             else // Increment distinct element count             {                int count = hM.get(arr[i]);                hM.put(arr[i], count+1);             }              // Print count of current window             System.out.println(dist_count);         }     }       // Driver method     public static void main(String arg[])     {         int arr[] =  {1, 2, 1, 3, 4, 2, 3};         int k = 4;         countDistinct(arr, k);     } }

Output:

3
4
4
3

Time complexity of the above solution is O(n).