Count Subarrays With Score Less Than K | OA LEVEL | TRICKY | MUST DO

https://leetcode.com/problems/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 <= 10^5
1 <= nums[i] <= 10^5
1 <= k <= 10^15


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class Solution {
public:
    long long countSubarrays(vector<int>& nums, long long k) {
        long long res=0;
        int wS=0;
        long long sum=0;
        for(int wE=0;wE<nums.size();wE++){
            sum+=nums[wE];
            while(sum*(wE-wS+1)>=k){
                sum-=nums[wS++];
            }
            res+=(wE-wS+1);
        }
        return res;
    }
};

Can you do this using Binary Search??