To solve Minimum Add to Make Parentheses Valid, track unmatched open and close parentheses while scanning the string. Each imbalance indicates a required insertion. By maintaining a counter rather than a full stack, you can efficiently compute the minimum moves in a single pass, reflecting the stack-based state management pattern intrinsic to this problem.
Problem Statement
You are given a string s consisting only of '(' and ')'. A string is considered valid if every opening parenthesis has a corresponding closing parenthesis and the pairs are properly nested. You may insert parentheses at any position in s to achieve validity.
Return the minimum number of parentheses insertions required to make s valid. For example, for s = "())", the minimum insertions is 1, and for s = "(((", the minimum insertions is 3. Optimize for a linear scan using stack-like state management.
Examples
Example 1
Input: s = "())"
Output: 1
Example details omitted.
Example 2
Input: s = "((("
Output: 3
Example details omitted.
Constraints
- 1 <= s.length <= 1000
- s[i] is either '(' or ')'.
Solution Approach
Use a counter for unmatched opens
Iterate through s, incrementing an open counter for '(' and decrementing for ')'. When a ')' occurs without a matching '(', increment a separate insertion counter. This simulates a stack while keeping space O(1).
Calculate final required insertions
After scanning the string, any remaining open parentheses require corresponding ')' insertions. The total minimum insertions is the sum of unmatched closes counted during iteration and remaining opens at the end.
Optimize with Greedy single pass
By greedily inserting whenever an imbalance is detected, you avoid maintaining a full stack. This single-pass strategy aligns with the Stack-based state management pattern and ensures linear time performance.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | O(N) |
| Space | O(1) |
Time complexity is O(N) as each character is visited once. Space complexity is O(1) because only counters are maintained instead of an actual stack, making it optimal for large strings.
What Interviewers Usually Probe
- Check if candidate handles both excess '(' and ')' correctly.
- Listen for discussion of O(1) space using counters versus explicit stack.
- Evaluate understanding of stack-based state tracking applied greedily.
Common Pitfalls or Variants
Common pitfalls
- Failing to count unmatched ')' during iteration leading to undercounted insertions.
- Using a full stack unnecessarily, increasing space usage.
- Ignoring leftover '(' at the end of the string.
Follow-up variants
- Compute minimum removals instead of insertions to make a parentheses string valid.
- Apply the same stack-based method to strings containing other types of brackets.
- Return the corrected valid string after minimal insertions rather than just the count.
How GhostInterview Helps
- Highlights which parentheses imbalances require insertions during iteration.
- Tracks unmatched opens and closes efficiently to reveal minimum operations.
- Simulates stack-based state without extra memory to improve runtime and clarity.
Topic Pages
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- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What pattern does Minimum Add to Make Parentheses Valid follow?
It follows the Stack-based state management pattern, where tracking unmatched parentheses determines minimum insertions.
Can this problem be solved without an explicit stack?
Yes, using counters for unmatched '(' and ')' achieves the same result with O(1) space.
Why is a single pass sufficient for this problem?
Each character's contribution to imbalance is independent, so tracking counters while scanning once captures all required insertions.
What common mistakes should be avoided?
Failing to account for leftover '(' at the end, ignoring unmatched ')', or unnecessarily using a full stack.
How does the greedy approach work here?
It inserts when an imbalance occurs rather than revisiting previous characters, ensuring minimum insertions efficiently.
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Open problem page