Subarray Sum Equals K with Bounded Window

Problem Context

You are given an integer array nums and two integers k and maxLen. Your task is to find the total number of contiguous subarrays whose sum equals k AND whose length does not exceed maxLen.

This problem combines the classic Two Sum / prefix-sum hashing technique with a sliding window / length constraint, making it a natural extension of common interview problems.

Input Format

• nums — a list of integers (can include negatives)
• k — the target subarray sum
• maxLen — the maximum allowed subarray length (inclusive)

Output Format

Return a single integer: the count of valid subarrays.

Examples

Example 1

Input: nums = [1, 2, 3, -2, 2], k = 3, maxLen = 3 Output: 3 Explanation:

• [1, 2] → sum = 3, length = 2 ✅
• [3] → sum = 3, length = 1 ✅
• [1, 2, 3, -2, -1] ← too long; skip
• [3, -2, 2] → sum = 3, length = 3 ✅ Valid subarrays: 3

Example 2

Input: nums = [1, -1, 1, -1], k = 0, maxLen = 2 Output: 3 Explanation:

• [1, -1] at indices 0–1 ✅
• [-1, 1] at indices 1–2 ✅
• [1, -1] at indices 2–3 ✅

Example 3

Input: nums = [5], k = 5, maxLen = 1 Output: 1 Explanation: The single element [5] equals k and has length 1 ≤ maxLen.

Approach Hints

1. Prefix Sum: Compute a running prefix sum prefixSum[i] = sum of nums[0..i-1]. A subarray nums[l..r] has sum = prefixSum[r+1] - prefixSum[l].
2. Hash Map: To find pairs where the difference equals k, store previously seen prefix sums. For each index r, look up how many times prefixSum[r] - k has appeared.
3. Length Constraint: The subarray nums[l..r] has length r - l + 1. You need r - l + 1 <= maxLen, i.e., l >= r - maxLen + 1. This means you cannot use all previous prefix sums — only those within the valid window.