Snapchat·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jul 2026

Summary

Snapchat software engineer interview that was pretty much all binary search variations. They pushed hard on both iterative and recursive implementations plus complexity analysis, which I wasn't fully expecting.

Questions Asked (1)

Given a sorted integer array and a target value, return the index of the first element that is greater than or equal to the target. Implement it both iteratively and recursively, handle duplicates, analyze the time and space complexity, and then explain how you'd modify the approach to find the last element less than or equal to the target.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The iterative part was fine, lower bound binary search, nothing too bad.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and edge cases, then implement iterative binary search to find the lower bound (first element >= target). Next, implement a recursive version, discuss time/space complexity, and finally explain how to adapt the logic to find the upper bound (last element <= target).

Pro tip: Emphasize that the iterative solution uses O(1) space while the recursive uses O(log n) due to call stack, and mention that in production you'd prefer the iterative version for efficiency. Also, note that the same binary search template can be easily modified for the upper bound by changing the comparison and update rules.

1. Clarify and define

Restate the problem: find the first index where array[index] >= target. Discuss edge cases: empty array, all elements less than target, all elements greater than or equal to target, duplicates.

2. Iterative implementation

Implement binary search iteratively using low and high pointers. Maintain the invariant that the answer is in [low, high]. When array[mid] >= target, update high = mid; else low = mid + 1. Return low after loop.

3. Recursive implementation

Implement the same logic recursively with a helper function that takes low and high. Base case: low == high, return low. Recursive case: if array[mid] >= target, recurse on [low, mid]; else recurse on [mid+1, high].

4. Complexity analysis

Time complexity: O(log n) for both iterative and recursive. Space complexity: O(1) for iterative, O(log n) for recursive due to call stack. Mention that recursion depth is logarithmic.

5. Modify for last element <= target

To find the last element <= target, change the condition: when array[mid] <= target, update low = mid; else high = mid - 1. Be careful with infinite loops; use mid = (low + high + 1) // 2 to avoid rounding down.

Key Points to Mention

Binary search for lower bound (first occurrence) and upper bound (last occurrence).
Handling duplicates: the algorithm naturally finds the first occurrence when using the correct update rules.
Edge cases: empty array, target smaller than all elements, target larger than all elements.
Time complexity O(log n) and space complexity O(1) iterative vs O(log n) recursive.
Modification for last element <= target: adjust comparison and midpoint calculation to avoid infinite loops.
Potential off-by-one errors and how to avoid them by maintaining a consistent invariant.

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