Leetcode #2302: Count Subarrays With Score Less Than K

In this guide, we solve Leetcode #2302 Count Subarrays With Score Less Than K in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

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.

Quick Facts

Difficulty: Hard
Premium: No
Tags: Array, Binary Search, Prefix Sum, Sliding Window

Intuition

We are looking for a contiguous region that satisfies a constraint, which is a classic sliding-window signal.

Expanding and shrinking the window lets us maintain validity without restarting the scan.

Approach

Grow the window with a right pointer, and shrink from the left only when the constraint is violated.

Track the best window as you go to keep the solution linear.

Steps:

Expand the right end of the window.
While invalid, move the left end to restore constraints.
Update the best window found.

Example

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.

Python Solution

class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        s = list(accumulate(nums, initial=0))
        ans = 0
        for i in range(1, len(s)):
            l, r = 0, i
            while l < r:
                mid = (l + r + 1) >> 1
                if (s[i] - s[i - mid]) * mid < k:
                    l = mid
                else:
                    r = mid - 1
            ans += l
        return ans

Complexity

The time complexity is $O(n \times \log n)$ , and the space complexity is $O(n)$ , where $n$ is the length of the array $\textit{nums}$ . The space complexity is $O(n)$ , where $n$ is the length of the array $\textit{nums}$ .

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

Example

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.

Python Solution

class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        s = list(accumulate(nums, initial=0))
        ans = 0
        for i in range(1, len(s)):
            l, r = 0, i
            while l < r:
                mid = (l + r + 1) >> 1
                if (s[i] - s[i - mid]) * mid < k:
                    l = mid
                else:
                    r = mid - 1
            ans += l
        return ans

Leetcode #2302: Count Subarrays With Score Less Than K

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #2302: Count Subarrays With Score Less Than K

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview