Tesla·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

May 2026

Summary

Tesla SWE interview that leaned heavily on classic algorithm problems. The trapping rainwater question came up and they wanted more than just a working solution, they wanted the full progression from naive to optimal.

Questions Asked (1)

Given an array of non-negative integers representing bar heights, calculate how much water gets trapped between the bars after rain. Walk through a brute-force solution first, analyze its complexity, then optimize to O(n) time and O(1) space.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

This is the classic trapping rainwater problem and I knew it, which was both good and bad.

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AI HintsAI Generated

Suggested Approach

Start by clearly restating the problem and confirming assumptions (e.g., non-negative heights, width of each bar is 1). Then present a brute-force solution that computes trapped water for each bar by finding the maximum height to its left and right, analyze its O(n^2) time complexity, and finally derive an O(n) time and O(1) space solution using two pointers with running maxima.

Pro tip: Emphasize that the two-pointer approach works because the water trapped at any position is determined by the minimum of the maximum heights on both sides; by moving the pointer with the smaller maximum, you ensure that the other side's maximum is already sufficient to bound the water. This demonstrates deep understanding and often impresses interviewers.

1. Clarify and Restate

Confirm the problem details: array of non-negative integers, each bar has width 1, and water is trapped between bars. Ask if there are any constraints or edge cases to consider.

2. Brute-Force Solution

For each bar, find the maximum height to its left and right, then add min(left_max, right_max) - height[i] to the total. Explain that this is O(n^2) time and O(1) space.

3. Optimize with Precomputed Arrays

Precompute left_max and right_max arrays in O(n) time and O(n) space, then compute trapped water in a single pass. This improves time to O(n) but uses extra space.

4. Two-Pointer O(1) Space Solution

Use two pointers (left and right) and maintain left_max and right_max. Move the pointer with the smaller max inward, adding water based on the current max. This achieves O(n) time and O(1) space.

5. Test and Discuss Trade-offs

Walk through a small example to verify correctness. Discuss trade-offs between the approaches, highlighting why the two-pointer method is optimal for space and time.

Key Points to Mention

The amount of water trapped at each bar is min(max_left, max_right) - height[i].
Brute-force time complexity is O(n^2) due to nested loops for finding max on each side.
Precomputed arrays reduce time to O(n) but increase space to O(n).
Two-pointer technique achieves O(n) time and O(1) space by maintaining running maxima.
Edge cases: empty array, all bars same height, strictly increasing/decreasing heights.
The two-pointer approach works because water is bounded by the smaller of the two maxima.

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