Bytedance·Data Scientist·Technical Phone Screen·Intermediate

Intermediate

Jul 2026

Summary

Bytedance Data Scientist interview with a coding problem on parentheses balancing. Pretty standard algorithmic screen but the problem had a small wrinkle that tripped me up at first.

Questions Asked (1)

Given a string of only '(' and ')' characters, what is the minimum number of parentheses you need to insert to make the string valid? Walk through your approach and its time and space complexity.

Algorithms & Data Structures

Author's notes

My first instinct was to just count unmatched brackets with a stack, which got me partway there.

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AI HintsAI Generated

Suggested Approach

Use a single-pass greedy algorithm with a counter for unmatched opening parentheses. Traverse the string, incrementing for '(' and decrementing for ')' when possible; count insertions needed for unmatched ')' and leftover '('.

Pro tip: Mention that this is equivalent to finding the minimum insertions to balance parentheses, and that the same logic can be applied to other bracket types with a stack. Also, note that the problem can be solved in O(n) time and O(1) space, which is optimal.

1. Clarify the problem

Confirm that the goal is to insert parentheses anywhere to make the string valid, and that we want the minimum number. Define what a valid parentheses string is.

2. Design the algorithm

Use a counter for unmatched opening parentheses. Iterate through the string: if '(', increment; if ')', decrement if counter > 0, else increment a counter for insertions needed for unmatched closing parentheses.

3. Compute the result

After the loop, the total insertions needed is the sum of insertions for unmatched closing parentheses and the remaining unmatched opening parentheses (the counter value).

4. Analyze complexity

Time complexity is O(n) because we traverse the string once. Space complexity is O(1) because we only use a few integer variables.

5. Test with examples

Walk through examples like '())', '(((', and '()' to verify the algorithm and edge cases.

Key Points to Mention

Greedy approach: always match closing parentheses with available opening ones.
Two counters: one for unmatched opening parentheses, one for insertions needed for unmatched closing parentheses.
Time complexity O(n) and space complexity O(1).
Edge cases: empty string, all opening, all closing, already valid.
Alternative stack-based approach with O(n) space, but not optimal.
The problem is equivalent to finding the minimum number of parentheses to add to balance.

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