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2302. Count Subarrays With Score Less Than K (Hard)

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

Companies: Amazon, Google

Related Topics:
Array, Binary Search, Sliding Window, Prefix Sum

Similar Questions:

• Maximum Subarray (Medium)
• Subarray Product Less Than K (Medium)
• Binary Subarrays With Sum (Medium)

Solution 1. Two Pointers

// OJ: https://leetcode.com/problems/count-subarrays-with-score-less-than-k
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    long long countSubarrays(vector<int>& A, long long k) {
        long long N = A.size(), ans = 0;
        vector<long long> sum(N + 1);
        for (int i = 0; i < N; ++i) sum[i + 1] = sum[i] + A[i];
        for (int i = 0, j = 1; j <= N; ++j) {
            while ((sum[j] - sum[i]) * (j - i) >= k) ++i;
            ans += j - i;
        }
        return ans;
    }
};

Solution 2. Sliding Window

// OJ: https://leetcode.com/problems/count-subarrays-with-score-less-than-k
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    long long countSubarrays(vector<int>& A, long long k) {
        long long N = A.size(), ans = 0, sum = 0;
        for (int i = 0, j = 0; j < N; ++j) {
            sum += A[j];
            while (sum * (j - i + 1) >= k) sum -= A[i++];
            ans += j - i + 1;
        }
        return ans;
    }
};