Went with prefix sums and a hashmap, which is the right move.
Start by clarifying the problem constraints (e.g., array size, element ranges, negative numbers) and then present a brute-force O(n^2) solution as a baseline. Optimize to O(n) using a hash map that stores prefix sum frequencies, explaining how it counts subarrays summing to k in a single pass. Walk through a small example to demonstrate correctness and discuss edge cases.
Pro tip: Mention that the hash map approach handles negative numbers and zeros seamlessly, unlike sliding window, and emphasize that initializing the map with {0: 1} is crucial for subarrays starting at index 0.
Ask about array size, element ranges (negative, zero, positive), and whether k can be negative. Discuss edge cases like empty array, single element, and large inputs.
Describe the O(n^2) approach: iterate over all subarrays, compute sum, and count those equal to k. Mention its time and space complexity.
Explain that for each index j, we need the number of indices i < j where prefix_sum[j] - prefix_sum[i] = k, i.e., prefix_sum[i] = prefix_sum[j] - k. Use a hash map to store frequencies of prefix sums seen so far.
Choose a small array (e.g., [1,2,3], k=3) and manually trace the algorithm, showing how the hash map updates and the count increments.
State that the optimized solution runs in O(n) time and O(n) space. Compare with brute-force and mention that the hash map approach is optimal for this problem.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a modified binary search that compares the middle element with its neighbors to determine which half contains a local minimum. If the middle is smaller than both neighbors, return it; otherwise, move toward the smaller neighbor. This guarantees O(log n) time because the search space halves each step.
Pro tip: Explicitly handle edge cases (single element, boundaries) and explain why the algorithm terminates—this shows you understand the invariant that a local minimum always exists in the chosen half.
Confirm the array is non-empty and adjacent elements are distinct. Handle single-element and boundary cases (index 0 or n-1) by checking if the first or last element is smaller than its only neighbor.
Initialize low = 0 and high = n-1. While low <= high, compute mid = low + (high - low) / 2.
Compare arr[mid] with its neighbors (if they exist). If arr[mid] is smaller than both, return mid.
If the left neighbor is smaller, set high = mid - 1; otherwise, set low = mid + 1. This moves toward a guaranteed local minimum.
Explain that each step halves the search space, giving O(log n) time and O(1) space. Justify why a local minimum must exist in the chosen half.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.