Subarrays with K Different Integers

import java.util.HashMap;

class Solution {
    public int subarraysWithKDistinct(int[] nums, int k) {
	    /*
			 * atMostKDistinct(nums, k):
			 *  Counts the number of subarrays with at most k distinct integers.
			 *  This includes subarrays with 0, 1, 2, ..., up to k distinct integers.
			 *
			 * atMostKDistinct(nums, k-1):
			 *  Counts the number of subarrays with at most k-1 distinct integers.
			 *  This includes subarrays with 0, 1, 2, ..., up to k-1 distinct integers.
			 ----------------------------------------------------------------------------------------------------
			 * exactlyKDistinct(nums, k) = atMostKDistinct(nums, k) - atMostKDistinct(nums, k-1)
		 */
        return atMostKDistinct(nums, k) - atMostKDistinct(nums, k - 1);
    }

    private int atMostKDistinct(int[] nums, int k) {
        HashMap<Integer, Integer> fMap = new HashMap<>();
        int start = 0;
        int end = 0;
        int count = 0;
        /*
         * 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 `end – start + 1`
         * where right and left are the boundaries of the current window.
         */
        while (end < nums.length) {
            // Add the current number to the frequency map
            fMap.put(nums[end], fMap.getOrDefault(nums[end], 0) + 1);

            // If the map size exceeds k, we need to shrink the window
            while (fMap.size() > k) {
                fMap.put(nums[start], fMap.get(nums[start]) - 1);
                if (fMap.get(nums[start]) == 0) {
                    fMap.remove(nums[start]);
                }
                start++;
            }

            // The number of subarrays with at most k distinct elements
            // ending at the current 'end' index is (end - start + 1)
            if (fMap.size() <= k) { // btw we don't need this condition, it's just for clarity
                int windowSize = end - start + 1;
                count += windowSize;
            }

            // Move the end pointer to the right
            end++;
        }

        return count;
    }
}