Count Subarrays With Score Less Than K

Problem

The score of an array is defined as the product of its sum and its length.

• For example, the score of [1, 2, 3, 4, 5] is (1 + 2 + 3 + 4 + 5) * 5 = 75.

Given a positive integer array nums and an integer k, return the number of non-empty subarrays of nums whose score is strictly less than k.

A subarray is a contiguous sequence of elements within an array.

Example 1:

Input: nums = [2,1,4,3,5], k = 10
Output: 6
Explanation:
The 6 subarrays having scores less than 10 are:

- [2] with score 2 * 1 = 2.
- [1] with score 1 * 1 = 1.
- [4] with score 4 * 1 = 4.
- [3] with score 3 * 1 = 3.
- [5] with score 5 * 1 = 5.
- [2,1] with score (2 + 1) * 2 = 6.
Note that subarrays such as [1,4] and [4,3,5] are not considered because their scores are 10 and 36 respectively, while we need scores strictly less than 10.

Example 2:

Input: nums = [1,1,1], k = 5
Output: 5
Explanation:
Every subarray except [1,1,1] has a score less than 5.
[1,1,1] has a score (1 + 1 + 1) * 3 = 9, which is greater than 5.
Thus, there are 5 subarrays having scores less than 5.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 105
• 1 <= k <= 1015

Solution

Note: this solution times out

class Solution {
    public long countSubarrays(int[] nums, long k) {
        var left = 0;
        var right = 0;
        var ans = 0;

        // we could pre-compute cumulative sum
        var sums = new int[nums.length];
        sums[0] = nums[0];
        for (var i = 1; i < nums.length; i++) {
            sums[i] = sums[i - 1] + nums[i];
        }

        while (left < nums.length) {
            var size = right - left + 1;
            var sum = sums[right] - sums[left] + nums[left];
            var score = sum * size;
            if (score < k) {
                ans += 1;
                if (right < nums.length - 1) {
                    right += 1;
                    continue;
                }
            }

            left += 1;
            right = left;
        }

        return ans;
    }
}