To solve this, traverse the string while using a stack to remove adjacent matching characters. This stack-based approach efficiently handles string state management. The result is the string that remains after no further removals can be made.
Problem Statement
You are given a string s consisting of lowercase English letters. Repeatedly perform the following operation: remove any two consecutive characters in the string that are the same.
Return the resulting string after no more such operations can be performed. The operation stops when no more adjacent matching characters exist.
Examples
Example 1
Input: s = "abc"
Output: "c"
Example 2
Input: s = "adcb"
Output: ""
Example 3
Input: s = "zadb"
Output: "db"
Constraints
- 1 <= s.length <= 105
- s consists only of lowercase English letters.
Solution Approach
Use a Stack for Character Removal
Traverse the string from left to right, pushing each character onto a stack. If the current character matches the one at the top of the stack, pop the stack to simulate removal. If not, push the character onto the stack. At the end of the traversal, the stack contains the resulting string.
Optimize for Performance
Since the string length can be large (up to 10^5), this approach ensures that each character is processed once, resulting in a time complexity of O(n). The space complexity is O(n) as well, due to the stack storing the characters.
Edge Cases Consideration
Handle edge cases such as strings with no adjacent characters to remove, strings with alternating characters, and strings that completely reduce to an empty string. Ensure the solution works for strings of varying lengths within the input constraints.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity of this solution is O(n), where n is the length of the string. Each character is pushed to and popped from the stack at most once. The space complexity is O(n) as the stack may hold up to n characters in the worst case.
What Interviewers Usually Probe
- Efficient handling of large strings with minimal overhead.
- Understanding of stack-based algorithms.
- Ability to identify and address edge cases in string manipulation.
Common Pitfalls or Variants
Common pitfalls
- Forgetting to handle cases where no characters can be removed.
- Misunderstanding the stack's behavior when removing characters.
- Failing to optimize for performance with large input sizes.
Follow-up variants
- Consider implementing this approach using a different data structure, such as a queue, to explore alternative solutions.
- Implement a variant that handles strings with special characters or case sensitivity.
- Modify the problem to allow multiple removals of different types of adjacent characters.
How GhostInterview Helps
- GhostInterview guides you through the stack-based approach, ensuring you understand how to traverse and manipulate the string effectively.
- The solver assists in breaking down edge cases and performance considerations, helping you anticipate real interview questions.
- You can practice optimizing space and time complexity within the constraints, a key part of problem-solving.
Topic Pages
Related GhostInterview Pages
- 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 is the time complexity of this approach?
The time complexity is O(n), where n is the length of the string, as each character is processed once.
How does the stack help in solving the problem?
The stack allows for efficient management of the string state, by pushing characters and removing them when adjacent matching ones are found.
What happens if the string has no adjacent matching characters?
If no adjacent matching characters exist, the stack will simply contain the original string as is, with no changes made.
What are the edge cases for this problem?
Edge cases include strings that are already reduced, strings with alternating characters, and strings that result in an empty string after removals.
How can this solution be optimized for large inputs?
The solution is already optimized with O(n) time and space complexity, ensuring efficient processing even for strings with lengths up to 10^5.
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Calculate the mirror score of a string using stack-based state management for matching letters efficiently and accurately.
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