LeetCode #2302 — HARD

Count Subarrays With Score Less Than K

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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.

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Sliding Window

Example 1

[2,1,4,3,5]
10

Example 2

[1,1,1]
5

Core Insight

What unlocks the optimal approach

If we add an element to a list of elements, how will the score change?
How can we use this to determine the number of subarrays with score less than k in a given range?
How can we use “Two Pointers” to generalize the solution, and thus count all possible subarrays?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2302: Count Subarrays With Score Less Than K
class Solution {
    public long countSubarrays(int[] nums, long k) {
        int n = nums.length;
        long[] s = new long[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        long ans = 0;
        for (int i = 1; i <= n; ++i) {
            int l = 0, r = i;
            while (l < r) {
                int mid = (l + r + 1) >> 1;
                if ((s[i] - s[i - mid]) * mid < k) {
                    l = mid;
                } else {
                    r = mid - 1;
                }
            }
            ans += l;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2302: Count Subarrays With Score Less Than K
func countSubarrays(nums []int, k int64) (ans int64) {
	n := len(nums)
	s := make([]int64, n+1)
	for i, x := range nums {
		s[i+1] = s[i] + int64(x)
	}
	for i := 1; i <= n; i++ {
		l, r := 0, i
		for l < r {
			mid := (l + r + 1) >> 1
			if (s[i]-s[i-mid])*int64(mid) < k {
				l = mid
			} else {
				r = mid - 1
			}
		}
		ans += int64(l)
	}
	return
}

# Accepted solution for LeetCode #2302: Count Subarrays With Score Less Than K
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

// Accepted solution for LeetCode #2302: Count Subarrays With Score Less Than K
impl Solution {
    pub fn count_subarrays(nums: Vec<i32>, k: i64) -> i64 {
        let n = nums.len();
        let mut s = vec![0i64; n + 1];
        for i in 0..n {
            s[i + 1] = s[i] + nums[i] as i64;
        }
        let mut ans = 0i64;
        for i in 1..=n {
            let mut l = 0;
            let mut r = i;
            while l < r {
                let mid = (l + r + 1) / 2;
                let sum = s[i] - s[i - mid];
                if sum * (mid as i64) < k {
                    l = mid;
                } else {
                    r = mid - 1;
                }
            }
            ans += l as i64;
        }
        ans
    }
}

// Accepted solution for LeetCode #2302: Count Subarrays With Score Less Than K
function countSubarrays(nums: number[], k: number): number {
    const n = nums.length;
    const s: number[] = Array(n + 1).fill(0);
    for (let i = 0; i < n; ++i) {
        s[i + 1] = s[i] + nums[i];
    }
    let ans = 0;
    for (let i = 1; i <= n; ++i) {
        let [l, r] = [0, i];
        while (l < r) {
            const mid = (l + r + 1) >> 1;
            if ((s[i] - s[i - mid]) * mid < k) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        ans += l;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.