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 thenumber of non-empty subarrays of nums whose score isstrictly less than k.

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

Examples

Example 1

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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

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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

Solution

Method 1 - Using Sliding Window

To solve this problem, we need to find the number of non-empty subarrays whose score (product of sum and length) is strictly less than k. Here’s the detailed explanation:

Approach

Sliding Window Technique:
- We use the sliding window method to efficiently compute the sum and length of subarrays in linear time. This avoids computing the score for all possible subarrays, which would take O(n^2) time.
Two Pointers:
- Maintain two pointers (l and r) to track the current subarray.
- Expand the window (r) to include more elements and add them to the cumulative sum.
- If the score (current sum * (r - l + 1)) exceeds k, increment l (contracting the window) until the score becomes less than k.
Count Valid Subarrays:
- For every r, the number of valid subarrays ending at r is (r - l + 1) because all subarrays starting from l to r satisfy the condition.

Code

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class Solution {

    public long countSubarrays(int[] nums, long k) {
        long ans = 0;
        long currSum = 0;
        int l = 0; // Left pointer

        for (int r = 0; r < nums.length; r++) { // Right pointer
            currSum += nums[r];

            // While the score of the subarray [l...r] is >= k, contract the window
            while (currSum * (r - l + 1) >= k) {
                currSum -= nums[l];
                l++;
            }

            // Add the number of valid subarrays ending at r
            ans += (r - l + 1);
        }

        return ans;
    }
}

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class Solution:
    def count_subarrays(self, nums: List[int], k: int) -> int:
        n = len(nums)
        ans = 0
        l = 0  # Left pointer
        curr_sum = 0

        for r in range(n):  # Right pointer
            curr_sum += nums[r]
            
            while curr_sum * (r - l + 1) >= k:
                curr_sum -= nums[l]
                l += 1
            
            ans += (r - l + 1)
        
        return ans

Complexity

⏰ Time complexity: O(n). The left pointer moves at most once for every right pointer. Thus, sliding window approach ensures O(n) time complexity.
🧺 Space complexity: O(1) as no additional space is used beyond variables for the algorithm.

Problem#

Examples#

Example 1#

Example 2#

Constraints#

Solution#

Method 1 - Using Sliding Window#

Approach#

Code#

Complexity#