Find power-of-two subarrays within sum range

Given an integer array nums and an integer k, find all subarrays whose length is a power of two and whose sum lies in the inclusive range [k, 2k]. Return the list of index ranges. Provide an algorithm faster than O(n^ 2). For non-negative nums, use sliding windows across candidate lengths (1, 2, 4, ...). For arrays with negative values, propose a prefix-sum plus balanced tree or hash-bucketing strategy and analyze complexity.

What a Strong Answer Covers

• A clear algorithm with the right data structures and enough pseudocode or code-level detail to implement it.
• A correctness argument that explains why the algorithm covers all required cases.
• Time and space complexity, plus at least one alternative approach when relevant.
• Focused tests for normal cases, edge cases, and failure modes.

Follow-up Questions

• How would the approach change if the input were streaming or too large for memory?
• What invariants would you assert in production code?
• Which tests would catch off-by-one, duplicate, or tie-breaking bugs?

Find power-of-two subarrays within sum range

Given an integer array nums and an integer k, find all subarrays whose length is a power of two (1, 2, 4, 8, ...) and whose sum lies in the inclusive range [k, 2k]. Return the list of index ranges as [start, end] pairs (both endpoints inclusive, 0-indexed).

Results must be returned ordered first by subarray length ascending (1, then 2, then 4, ...), and within the same length by start index ascending.

Approach. A brute-force scan over all O(n) start positions and all O(log n) candidate power-of-two lengths, recomputing each window sum, costs O(n^2 log n). Precomputing a prefix-sum array lets each window sum be read in O(1), so the whole scan is O(n log n) — comfortably faster than O(n^2). The prefix-sum method works uniformly for non-negative and negative values (no sliding-window monotonicity assumption is needed), which is exactly why it is preferred when nums can contain negatives.

Examples.

• nums = [1, 2, 3, 4], k = 3 → [[2, 2], [3, 3], [0, 1], [1, 2]] (length-1 windows summing to 3 or 4, then length-2 windows summing in [3, 6]).
• nums = [4, -1, 2, 5], k = 3 → [[0, 0], [3, 3], [0, 1]] (negative values handled correctly).
• nums = [], k = 1 → [].

Constraints

• 0 <= len(nums) <= 10^5
• nums[i] may be negative, zero, or positive
• k may be negative, zero, or positive; the range [k, 2k] is empty when k < 0 (since then 2k < k)
• Only subarray lengths that are exact powers of two (1, 2, 4, 8, ...) are considered
• Index ranges are 0-indexed and inclusive on both ends

Hints

1. Precompute prefix sums so any window sum is prefix[start+length] - prefix[start] in O(1).
2. The only candidate lengths are powers of two; there are at most floor(log2(n)) + 1 of them, so iterate length = 1, 2, 4, ... up to n.
3. For each length, slide the start from 0 to n - length and test whether the window sum falls in [k, 2k]. This handles negative values naturally — no monotonic sliding-window trick is required.
4. Emit results ordered by length first, then by start index, to get a stable, predictable output.

Solution

Solution Alignment

The prompt asks for an implementation-level answer. The safest way to present it is to define the state, maintain clear invariants, then walk through complexity and tests.

Problem Restatement

Given an integer array nums and an integer k, find all subarrays whose length is a power of two and whose sum lies in the inclusive range [k, 2k]. Return the list of index ranges. Provide an algorithm faster than O(n^ 2). For non-negative nums, use sliding windows across candidate lengths (1, 2, 4, ...). For arrays with negative values, propose a prefix-sum plus balanced tree or hash-bucketing strategy and analyze complexity.

Recommended Approach

Choose traversal based on the required output. DFS is natural for subtree computations, reconstruction, and range pruning; BFS is natural for level order or side views. Keep per-depth or per-position state when the output depends on columns, rows, or depths.

Correctness

The implementation should maintain an invariant after each loop or operation that directly matches the problem statement. At termination, that invariant implies the returned value has considered every valid candidate exactly once, or has preserved the required data-structure state after every API call.

Complexity

Most tree traversals are O(n) time and O(h) recursion stack for DFS or O(w) queue space for BFS, where h is height and w is maximum width.

Edge Cases and Tests

Empty tree, one node, skewed tree, duplicate values when reconstruction assumes uniqueness, deep recursion, and tie-breaking for same row/column nodes.