2394 Count Subarrays With Score Less Than K

Count Subarrays With Score Less Than K

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

class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:

        left = 0
        s = sum(nums)
        total = 0
        res = 0
        right = 0

        while(left < len(nums)):
            while right < len(nums) and (total + nums[right]) * (right - left + 1) < k:
                total += nums[right]
                right += 1

            # After expansion, [left, right-1] are all valid
            res += right - left

            # Move left forward
            total -= nums[left]
            left += 1
        return res

        



        right = len(nums) - 1
        
        while(left < len(nums)):
            if s * (right - left + 1) >= k: # still large
                s -= nums[right]
                right -= 1
                if right < 0:
                    left += 1
                    s = left
                    right = left
            else:
                # found max < k

                res += right - left + 1 # nums of subarray
                s -= nums[left]
                left += 1
                while(s * (right - left + 1) <= k and right < len(nums)-1):
                    right += 1
                    s += nums[right]
        return res