Approach:

Iterate over the array from index 0 to n - k to consider all windows of size k.
For each window starting at index i, call windows(arr + i, k) which checks for distinct elements in that window.
Inside windows(), for each element in the window, compare it with all previous elements to check if it's a duplicate. If it's not found before, it's considered unique.
Print the count of distinct elements for every window.

Code

#include <bits/stdc++.h>
using namespace std;

// Function to count distinct elements in a given subarray of size k
int windows(int arr[], int k) {
    // Variable to store distinct count
    int count = 0;

    // Traverse each element in the current window
    for (int i = 0; i < k; i++) {
        int j;
        // Check if this element has appeared before in this window
        for (j = 0; j < i; j++) {
            if (arr[i] == arr[j])
                break;
        }
        // If element not found before, increment distinct count
        if (j == i)
            count++;
    }
    return count;
}

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