The power-of-two length constraint is what makes this interesting.
First, clarify that the subarray length must be a power of two, so only lengths 1, 2, 4, 8, ... up to n are valid. Use prefix sums to compute any subarray sum in O(1), then for each valid length L, slide a window of size L across the array and check if the sum is in [k, 2k]. This yields O(n log n) time, which is faster than O(n^2).
Pro tip: Mention that the number of power-of-two lengths up to n is O(log n), so the total work is O(n log n). Also note that if k is negative, the range [k, 2k] is empty (since k > 2k), so you can return an empty list immediately—this edge case shows attention to detail.
Confirm that subarray length must be a power of two (1, 2, 4, ...) and that the sum must be in [k, 2k]. Check if k is negative: if so, the range is invalid and you can return an empty result.
Build a prefix sum array where prefix[i] = sum of first i elements. This allows computing the sum of any subarray arr[l..r] in O(1) as prefix[r+1] - prefix[l].
For each power of two L from 1 up to n, slide a window of length L across the array. For each window, compute its sum using prefix sums and check if it lies in [k, 2k].
When a valid subarray is found, record its start and end indices (inclusive). Return the list of all such index ranges.
Explain that there are O(log n) valid lengths, and for each length we do O(n) work, giving O(n log n) time and O(n) space for prefix sums. Discuss potential optimizations or alternative approaches if needed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the problem: if it's about finding a subarray with sum zero or a target sum, negative values break sliding window. Then, explain that prefix sums transform the problem into finding two equal (or target-differing) prefix sums, which can be done efficiently with a hash map or balanced tree. Finally, analyze time complexity and trade-offs between the two data structures.
Pro tip: Mention that a hash map gives O(n) average time but O(n) space, while a balanced tree gives O(n log n) time and can handle range queries or ordered traversal if needed. This shows you consider both average and worst-case scenarios.
Ask whether the goal is to find a subarray with sum zero, a target sum, or something else. Confirm if the array is static or dynamic, and if we need to return indices or just a boolean.
Sliding window relies on monotonicity of prefix sums; with negatives, the sum can decrease, so the window cannot be adjusted greedily. Thus, we need a different approach.
Define prefix sum P[i] = sum of first i elements. A subarray sum from i+1 to j equals P[j] - P[i]. So finding a subarray with sum S is equivalent to finding indices i < j with P[j] - P[i] = S, i.e., P[i] = P[j] - S.
Use a hash map to store prefix sums and their earliest index for O(1) average lookup. Alternatively, use a balanced BST (e.g., TreeMap) to store prefix sums, allowing O(log n) lookup and enabling ordered queries if needed.
Hash map: O(n) average time, O(n) space; worst-case O(n^2) if many collisions. Balanced tree: O(n log n) time, O(n) space, with guaranteed performance. Discuss trade-offs based on expected input and requirements.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.