Tesla·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026

Summary

Tesla SWE interview that went deep on a classic algorithm problem. They wanted the full arc: brute force first, then walk up to linear time, then squeeze out the extra space. Not a vibe check, more like a whiteboard lecture you're giving to yourself.

Questions Asked (1)

Implement a solution to the Trapping Rain Water problem. Start with a brute-force approach, explain its time and space complexity, then optimize to O(n) time with O(n) space, and finally discuss how you'd get it down to O(1) space.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The brute force part was fine, nested loops, nothing to it.

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AI HintsAI Generated

Suggested Approach

Start by restating the problem and clarifying assumptions (e.g., elevation map is non-negative, width of each bar is 1). Then walk through the brute-force O(n^2) solution, explain its complexity, and progressively optimize to O(n) time with O(n) space using precomputed max arrays, and finally to O(1) space using two pointers. Emphasize the trade-offs and reasoning at each step.

Pro tip: At Tesla, interviewers value first-principles thinking and efficiency. Explicitly connect the optimization to real-world constraints like memory limits on embedded systems, and mention that the two-pointer approach is often preferred in production due to its constant space.

1. Clarify and Restate

Confirm the problem: given an array of non-negative integers representing elevation, compute total trapped water. Clarify that each bar has width 1 and water cannot be trapped outside the array.

2. Brute Force Approach

For each index, 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 to O(n) Time, O(n) Space

Precompute left_max and right_max arrays in two passes, then compute trapped water in a third pass. This reduces time to O(n) at the cost of O(n) extra space.

4. Optimize to O(1) Space

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

5. Discuss Trade-offs and Edge Cases

Compare the approaches: brute force is simple but slow; precomputed arrays are faster but use extra memory; two-pointer is optimal for space. Mention edge cases like empty array, single bar, or all bars of equal height.

Key Points to Mention

Time and space complexity of each approach: O(n^2) vs O(n) time, O(1) vs O(n) space.
The key insight: water trapped at index i is min(max_left, max_right) - height[i].
How precomputing left and right max arrays eliminates redundant calculations.
The two-pointer technique and why it correctly computes trapped water without extra space.
Trade-offs between time and space, and when each approach might be appropriate.
Edge cases: empty input, single element, strictly increasing/decreasing heights, and flat terrain.

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