The core logic clicks pretty fast once you think about it: if the element to the right of mid is larger, there has to be a peak somewhere in that direction, so you go right; otherwise you go left and include mid.
Use a binary search approach by comparing the middle element with its neighbors to determine which half contains a peak. If the middle element is greater than both neighbors, it's a peak; otherwise, move towards the side with the larger neighbor. This works because the array has no two adjacent equal elements, guaranteeing a peak exists and can be found in O(log n) time.
Pro tip: Emphasize that the algorithm relies on the property that a peak must exist and that by moving towards the larger neighbor, you are guaranteed to find a peak. This demonstrates understanding of the problem's invariants and the correctness of the binary search approach.
Confirm that out-of-bounds positions are treated as negative infinity, and that no two adjacent elements are equal. Discuss edge cases like arrays of size 1 or 2.
Explain that you'll use binary search to find a peak by comparing the middle element with its neighbors and deciding which half to search next.
If the middle element is greater than both neighbors, return its index. If the left neighbor is greater, search the left half; otherwise, search the right half.
State that the algorithm runs in O(log n) time due to halving the search space each iteration, and uses O(1) extra space.
Explain why the algorithm always finds a peak: because the array has no equal adjacent elements, and moving towards the larger neighbor guarantees a peak exists in that direction.
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Did not see this coming on what I thought was a phone screen level question.
Start by clarifying the problem: a peak is any element greater than its four orthogonal neighbors. Then present a binary search on columns (or rows) approach: pick the middle column, find the global maximum in that column, compare it with its left and right neighbors, and recurse on the half that contains a larger neighbor. This achieves O(n log m) time, which is optimal for this problem.
Pro tip: Mention that the 2D peak-finding problem is a classic divide-and-conquer algorithm often asked at Meta, and that the key insight is reducing the search space by half each step. Also, note that while a linear scan is O(nm), the binary search approach is more efficient and demonstrates algorithmic maturity.
Confirm that a peak is defined as an element strictly greater than its adjacent neighbors (up, down, left, right), and ask about matrix dimensions, possible duplicates, and whether any peak is acceptable.
Describe how to pick the middle column, find the maximum element in that column, and compare it with its left and right neighbors to decide which half to search next.
Detail the recursive or iterative process: if the max in the middle column is greater than both left and right neighbors, it's a peak; otherwise, move to the half with the larger neighbor and repeat.
State that the time complexity is O(n log m) where n is the number of rows and m is the number of columns (or vice versa), and space complexity is O(1) for iterative or O(log m) for recursive.
Mention handling of 1D arrays, single row/column matrices, and compare with the naive O(nm) approach to highlight efficiency gains.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.