Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

May 2026Remote

Summary

Meta coding screen, got a string manipulation problem that looked deceptively simple at first glance. The stack-based approach was the whole point and they wanted O(n) explicitly.

Questions Asked (1)

Given a string, repeatedly remove any group of two or more consecutive identical characters until no such groups remain. Return the final string. Solve it in a single pass using a stack of (char, count) pairs in O(n) time and O(n) space.

Algorithms & Data Structures

Author's notes

The naive approach is obvious: scan, remove runs, repeat.

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AI HintsAI Generated

Suggested Approach

Use a stack of (char, count) pairs to process the string in one pass. For each character, if it matches the top of the stack, increment the count; otherwise, push a new pair. After updating, if the count reaches 2 or more, pop the pair, as those characters would be removed. At the end, reconstruct the string from the remaining pairs.

Pro tip: Clarify that the removal is iterative and can cascade, but the stack approach handles cascades automatically because after popping, the new top may match the next character, effectively merging groups. This shows you understand the problem's depth beyond a naive simulation.

1. Understand the problem and constraints

Restate the problem: repeatedly remove groups of 2+ identical consecutive characters until no such groups remain. Note that removals can cause new groups to form. The solution must be O(n) time and O(n) space, so a single-pass stack approach is ideal.

2. Design the stack-based algorithm

Use a stack where each element is a pair (char, count). Iterate through the string: if the stack is empty or the current char differs from the top's char, push (char, 1). If it matches, increment the top's count. If the count becomes 2 or more, pop the pair.

3. Walk through an example

Trace the algorithm on a sample string like 'abbaca' to demonstrate how groups are removed and how cascading works. Show the stack state at each step to verify correctness.

4. Analyze complexity and edge cases

Explain that each character is processed once, and each stack operation is O(1), giving O(n) time. Space is O(n) in the worst case. Discuss edge cases: empty string, no removals, all characters removed, and large groups.

5. Implement and test

Write clean code with meaningful variable names. Test with the example and edge cases. If time permits, discuss alternative approaches (e.g., two-pointer) and why the stack is optimal.

Key Points to Mention

The stack stores (char, count) pairs to efficiently track consecutive identical characters.
When count reaches 2, pop the pair to simulate removal; this may cause the new top to match the next character, handling cascading removals.
Time complexity is O(n) because each character is pushed and popped at most once.
Space complexity is O(n) due to the stack, which in the worst case holds all characters.
Edge cases: empty string, string with no removals, string that reduces to empty, and strings with large groups.
The algorithm is a single pass, which is optimal for this problem.

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