← Instacart Interview Insights
The naive approach is obvious and they explicitly block it with the complexity requirement.
Use two binary searches: one to find the leftmost occurrence of the target and another to find the rightmost occurrence. Modify the standard binary search to continue searching even after finding the target, adjusting the search space based on whether you're looking for the first or last position. This ensures O(log n) time complexity.
Pro tip: Clarify edge cases upfront (e.g., empty array, target not present) and mention that you'll handle them explicitly. Also, discuss how you would test the solution with examples, showing thoroughness.
Confirm the input array is sorted and may contain duplicates. Discuss edge cases: empty array, target smaller than all elements, target larger than all elements, target not present.
Implement a modified binary search that finds the leftmost index of the target. When the target is found, record the index and continue searching in the left half to find an earlier occurrence.
Similarly, implement a modified binary search for the rightmost index. When the target is found, record the index and continue searching in the right half.
If either search fails to find the target, return [-1, -1]. Otherwise, return the first and last indices found.
State that both searches run in O(log n) time, so overall O(log n). Walk through a few test cases to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.