Bytedance·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Bytedance software engineer screen, one coding problem, pretty standard stuff but the O(log n) constraint makes it clear they want binary search and not just a linear scan.

Questions Asked (1)

Given a sorted array of distinct integers and a target value, return the index if the target exists, or the index where it would be inserted to keep the array sorted. Must run in O(log n).

Algorithms & Data Structures

Author's notes

The O(log n) requirement is basically them telling you to use binary search without saying it.

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AI HintsAI Generated

Suggested Approach

Use binary search to find the target or the insertion point in O(log n) time. Maintain low and high pointers, and at each step compare the middle element with the target, narrowing the search range accordingly. Return low when the search ends, as it indicates the correct insertion index.

Pro tip: Explicitly state that you're using binary search and that the return value is the left boundary (low) after the loop, which naturally handles both found and not-found cases. This shows you understand the invariant and avoids off-by-one errors.

1. Clarify the problem and constraints

Confirm that the array is sorted, contains distinct integers, and that O(log n) time is required. Ask if there are any edge cases to consider, such as empty array or target outside the range.

2. Choose binary search and define invariants

Decide to use binary search. Set low = 0 and high = n (or n-1 depending on implementation). Maintain the invariant that the answer lies in [low, high].

3. Implement the search loop

While low < high (or low <= high), compute mid = low + (high - low) // 2. If nums[mid] < target, set low = mid + 1; else set high = mid. This ensures low ends at the first index where nums[index] >= target.

4. Return the result

After the loop, return low. This index is either the position of the target if it exists, or the insertion point if it does not.

5. Test with edge cases

Walk through examples: target smaller than all elements, larger than all, present in middle, and empty array. Verify that the returned index maintains sorted order.

Key Points to Mention

Time complexity: O(log n) due to halving the search space each iteration.
Space complexity: O(1) as only a few variables are used.
Binary search variant: finding the leftmost insertion point (lower bound).
Handling of edge cases: empty array, target at boundaries, target not present.
Use of mid = low + (high - low) // 2 to avoid integer overflow.
Correctness argument: invariant that low is the smallest index where nums[index] >= target.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.