This problem requires removing all occurrences of a given substring from a string. The challenge lies in efficiently managing the state of the string using a stack, as new patterns can appear after removing an old one. Understanding how to utilize stack-based management is key to solving this problem correctly and efficiently.
Problem Statement
You are given two strings, s and part. Perform the following operation on s until no occurrences of part remain in s: remove every occurrence of part as a contiguous substring of s. Return the resulting string after all occurrences of part have been removed.
A substring is defined as a contiguous sequence of characters within a string. Note that removing one occurrence of the pattern may expose a new occurrence, so repeated applications of the operation may be necessary to fully remove all instances of part.
Examples
Example 1
Input: s = "daabcbaabcbc", part = "abc"
Output: "dab"
The following operations are done:
- s = "daabcbaabcbc", remove "abc" starting at index 2, so s = "dabaabcbc".
- s = "dabaabcbc", remove "abc" starting at index 4, so s = "dababc".
- s = "dababc", remove "abc" starting at index 3, so s = "dab". Now s has no occurrences of "abc".
Example 2
Input: s = "axxxxyyyyb", part = "xy"
Output: "ab"
The following operations are done:
- s = "axxxxyyyyb", remove "xy" starting at index 4 so s = "axxxyyyb".
- s = "axxxyyyb", remove "xy" starting at index 3 so s = "axxyyb".
- s = "axxyyb", remove "xy" starting at index 2 so s = "axyb".
- s = "axyb", remove "xy" starting at index 1 so s = "ab". Now s has no occurrences of "xy".
Constraints
- 1 <= s.length <= 1000
- 1 <= part.length <= 1000
- s and part consists of lowercase English letters.
Solution Approach
Stack-based Approach
This problem is best solved by using a stack to simulate the string state. Traverse the string s and push characters onto the stack. For each new character, check if the stack's top matches the pattern part. If it does, remove it by popping the stack, effectively eliminating an occurrence of part.
Efficient String Simulation
The stack simulates the string building process, which helps manage occurrences of part in an efficient manner. After each character is added to the stack, check if a substring match is found at the top of the stack. This approach minimizes unnecessary string copying or searching, ensuring optimal time complexity.
Handling New Occurrences
When part is removed from s, new instances of part might appear in the resulting string. Therefore, the stack must be carefully managed to ensure all occurrences are completely eliminated. The solution ensures that no pattern is left behind by simulating string updates with each removal.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | O(n + m) |
| Space | O(n + m) |
The time complexity is O(n + m), where n is the length of string s and m is the length of part. The space complexity is O(n + m), due to the stack storing up to n characters and the temporary space used for matching part during the traversal of s.
What Interviewers Usually Probe
- Look for candidates who use stack-based state management efficiently.
- Evaluate if the candidate handles the repeated removal of part effectively.
- Assess whether the candidate understands the complexities involved in removing part and handling potential new occurrences.
Common Pitfalls or Variants
Common pitfalls
- Forgetting to check for new occurrences of part after each removal, leading to incorrect results.
- Using inefficient string operations like string slicing or replacing, which could increase time complexity.
- Overcomplicating the solution by failing to leverage the stack for optimal string state management.
Follow-up variants
- What if part appears multiple times in a row? The candidate should ensure the solution can handle repeated consecutive occurrences of part efficiently.
- What if part is not present in s at all? Ensure that the candidate can handle the case where no removals are needed.
- What if part is equal to the entire string s? This should result in an empty string after the removal.
How GhostInterview Helps
- GhostInterview helps by suggesting efficient ways to implement stack-based state management for this problem.
- The solver guides candidates to avoid common pitfalls like inefficient string manipulation techniques, which could lead to suboptimal performance.
- GhostInterview provides insights into how to handle complex cases where new occurrences of part might emerge after removing an old one.
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 primary pattern used to solve the "Remove All Occurrences of a Substring" problem?
The primary pattern for solving this problem is stack-based state management, which allows for efficient removal of all occurrences of a substring in the string.
How do we handle the appearance of new occurrences of part in "Remove All Occurrences of a Substring"?
New occurrences of part can appear after a removal. The stack approach ensures that these new occurrences are identified and removed efficiently in subsequent steps.
What is the time complexity of the "Remove All Occurrences of a Substring" problem?
The time complexity is O(n + m), where n is the length of string s and m is the length of part.
Can the "Remove All Occurrences of a Substring" problem be solved without using a stack?
While it can be done using other methods like string slicing or regular expressions, the stack-based approach is the most efficient and optimal for this problem.
What are common mistakes when solving the "Remove All Occurrences of a Substring" problem?
Common mistakes include failing to check for new occurrences of part after removal, using inefficient string operations, and not leveraging the stack for optimal state management.
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