LeetCode 2302: Count Subarrays With Score Less Than K Solution

Master LeetCode problem 2302 (Count Subarrays With Score Less Than K), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2302. Count Subarrays With Score Less Than K

Hard

Array Binary Search Sliding Window Prefix Sum

Problem Explanation

Explanation:

To solve this problem, we can use a sliding window approach. We initialize two pointers, left and right, and move the right pointer to expand the window while keeping track of the sum and length of the current subarray. If the score of the current subarray is less than k, we count the number of subarrays that can be formed with the current right pointer position. If the score exceeds k, we move the left pointer to contract the window until the score is less than k again.

Algorithmic Idea:

Initialize pointers left and right to 0.
Initialize variables sum and count to 0.
Iterate over the array with the right pointer.
Update sum and count based on the score of the current subarray.
If the score exceeds k, move the left pointer to shrink the window.
Repeat steps 3-5 until the end of the array.

Time Complexity:

The time complexity of this solution is O(n), where n is the number of elements in the input array.

Space Complexity:

The space complexity is O(1) as we are using a constant amount of extra space.

Solution Code

class Solution {
    public int numSubarrayProductLessThanK(int[] nums, int k) {
        int n = nums.length;
        int left = 0, sum = 0, count = 0;
        
        for (int right = 0; right < n; right++) {
            sum += nums[right];
            while (sum * (right - left + 1) >= k && left <= right) {
                sum -= nums[left];
                left++;
            }
            count += right - left + 1;
        }
        
        return count;
    }
}

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