I went straight to the linear scan from the end, which works fine but is O(n) and I kind of knew they'd push back.
Start by clarifying the problem constraints and edge cases, then explain the linear scan solution as a baseline. Next, introduce binary search by leveraging the monotonic property of failures to achieve O(log n) time, and discuss trade-offs between the two approaches.
Pro tip: Emphasize that binary search is only valid because failures are monotonic; without that guarantee, you'd need a linear scan. Also, mention that in a real-world frontend context, you might debounce or batch logs to avoid performance bottlenecks.
Ask about input format, size, and whether the array is sorted by timestamp. Confirm that failures are monotonic and that you need the first failure's timestamp.
Iterate through the logs from start to end, returning the timestamp of the first failure encountered. If none, return a sentinel value like -1 or null.
Use binary search to find the leftmost failure. Maintain low and high pointers, check the middle element: if it's a failure, move high to mid; else move low to mid+1. After the loop, check if low is within bounds and points to a failure.
Discuss time and space complexity: linear scan is O(n) time and O(1) space; binary search is O(log n) time and O(1) space. Mention that binary search requires the monotonic property and random access, which arrays provide.
Consider empty array, all successes, all failures, and single-element array. Ensure the binary search correctly handles these cases.
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