Meta·Software Engineer·Onsite - Coding / Algorithms·Intermediate

IntermediatePrefer not to say

Jun 2026

Summary

Meta SWE coding round, got a classic two-pointer problem on trapping rainwater. Nothing too exotic but it's the kind of question where you either know the pattern or you don't.

Questions Asked (1)

Given an array of non-negative integers representing bar heights, compute the total amount of water that can be trapped between the bars after rain.

Algorithms & Data Structures

Author's notes

Two pointers is the move here.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and walking through a small example to ensure understanding. Then, discuss a brute-force approach and optimize it using precomputed left and right maximum arrays or a two-pointer technique. Finally, analyze time and space complexity and test with edge cases.

Pro tip: Mention that the two-pointer approach achieves O(n) time and O(1) space, which is optimal. Also, relate the problem to real-world scenarios like trapping rainwater between buildings to show practical insight.

1. Clarify and Example

Restate the problem in your own words and walk through a simple example to confirm understanding. Ask clarifying questions about input constraints and expected output.

2. Brute Force Approach

Explain a naive solution that for each bar, finds the maximum height on the left and right, and adds the minimum of those minus the current height. This takes O(n^2) time.

3. Optimized Approach

Describe how to precompute left and right maximum arrays in O(n) time and O(n) space, or use two pointers to achieve O(n) time and O(1) space. Explain the logic clearly.

4. Complexity Analysis

State the time and space complexity of your chosen approach. Compare with the brute force to highlight the improvement.

5. Test with Edge Cases

Walk through edge cases such as empty array, all bars of same height, strictly increasing/decreasing heights, and arrays with zeros. Verify the solution handles them correctly.

Key Points to Mention

The amount of water trapped at each index is determined by the minimum of the maximum height to its left and right, minus its own height.
Precomputing left and right maximum arrays allows O(n) time and O(n) space solution.
Two-pointer technique optimizes space to O(1) while maintaining O(n) time.
Edge cases: empty array, single bar, all bars same height, strictly increasing/decreasing heights.
Time and space complexity trade-offs between different approaches.
Real-world analogy: water trapped between buildings after rain.

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