Meta·Software Engineer·Onsite - Coding / Algorithms·Intermediate

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Jun 2026

Summary

Meta SWE coding round, one algorithmic question on finding a local minimum in an array. The constraint was O(log n) so binary search was the expected approach.

Questions Asked (1)

Given an integer array, find and return the index of any local minimum, where the element is strictly smaller than its neighbors. Boundary elements only need to be smaller than their single neighbor. The solution must run in O(log n) time.

Algorithms & Data Structures

Author's notes

My first instinct was a linear scan and I actually started explaining it before catching myself.

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AI HintsAI Generated

Suggested Approach

Use a binary search approach by comparing the middle element with its neighbors to determine which half contains a local minimum. If the middle element is greater than its right neighbor, a local minimum must exist in the right half; otherwise, it exists in the left half (including the middle). Continue narrowing down until you find a local minimum.

Pro tip: Explicitly handle edge cases (array of size 1 or 2) and clarify that the algorithm works because the array can be thought of as having virtual -∞ boundaries, guaranteeing a local minimum exists. This shows attention to detail and deep understanding.

1. Clarify the problem and constraints

Confirm that the array is non-empty, elements are distinct? (Not necessarily, but strict inequality means no equal neighbors). Discuss edge cases: single element, two elements, and boundaries.

2. Explain the binary search strategy

Describe how to use binary search: at each step, compare the middle element with its neighbors. If it's a local minimum, return its index. Otherwise, decide which half to search based on the slope.

3. Detail the decision logic

If mid > mid+1, then a local minimum exists in the right half (including mid+1). Else if mid > mid-1, then a local minimum exists in the left half (including mid-1). Otherwise, mid itself is a local minimum.

4. Handle boundaries and termination

Adjust comparisons for the first and last elements. Ensure the loop terminates when the search space reduces to one element, which must be a local minimum.

5. Analyze complexity and test

State that the time complexity is O(log n) due to halving the search space each iteration, and space complexity is O(1). Walk through a small example to verify correctness.

Key Points to Mention

Binary search is applicable because the array can be viewed as having a 'valley' somewhere due to virtual boundaries at -∞.
The algorithm guarantees finding a local minimum, not necessarily the global minimum.
Time complexity: O(log n) because we halve the search space each step.
Space complexity: O(1) as we only use a few variables.
Edge cases: array of size 1 (return 0), size 2 (return index of smaller element), and boundaries.
The solution works even if there are multiple local minima; we just need to find any one.

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