Count distinct elements in every window

Problem Statement - link #

Given an array of integers and a number K. Find the count of distinct elements in every window of size K in the array.

Your Task:

Your task is to complete the function countDistinct() which takes the array A[], the size of the array(N) and the window size(K) as inputs and returns an array containing the count of distinct elements in every contiguous window of size K in the array A[].

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

Examples #

Example 1:

Input:
N = 7, K = 4
A[] = {1,2,1,3,4,2,3}
Output: 3 4 4 3
Explanation: Window 1 of size k = 4 is
1 2 1 3. Number of distinct elements in
this window are 3. 
Window 2 of size k = 4 is 2 1 3 4. Number
of distinct elements in this window are 4.
Window 3 of size k = 4 is 1 3 4 2. Number
of distinct elements in this window are 4.
Window 4 of size k = 4 is 3 4 2 3. Number
of distinct elements in this window are 3.

Example 2:

Input:
N = 3, K = 2
A[] = {4,1,1}
Output: 2 1

Constraints #

• 1 <= K <= N <= 10^5
• 1 <= A[i] <= 10^5

Solutions #

class Solution {
  public:
    vector <int> countDistinct (int A[], int n, int k) {
        //code here.
        unordered_map<int, int> mp;
        for(int i = 0; i < k; i++) {
            mp[A[i]]++;
        }
        int j = k;
        vector<int> ans;
        while(j < n) {
            ans.push_back(mp.size());
            if(mp[A[j-k]] > 1) {
                mp[A[j-k]]--;
            }
            else {
                mp.erase(A[j-k]);
            }
            mp[A[j]]++;
            j++;
        }
        ans.push_back(mp.size());
        return ans;
    }
};