#2390

Medium

auto_awesomeStack-based state management

LeetCode Problem Workspace

Removing Stars From a String

Remove all stars from a string by simulating the removal process using a stack to manage characters and stars efficiently.

string stack simulation

Problem Statement

You are given a string s that may contain stars (*). In each operation, you must remove the most recently added character when encountering a star and continue until all stars are processed. Return the string after all stars have been removed.

The string consists of lowercase English letters and stars, and the task is to simulate the process efficiently. The challenge emphasizes managing characters with a stack, ensuring the correct output even as stars trigger removals from the left side of the string.

Examples

Example 1

Input: s = "leetcod*e"

Output: "lecoe"

Performing the removals from left to right:

The closest character to the 1st star is 't' in "leetcod e". s becomes "lee cod*e".
The closest character to the 2nd star is 'e' in "lee cod e". s becomes "lecod*e".
The closest character to the 3rd star is 'd' in "lecod*e". s becomes "lecoe". There are no more stars, so we return "lecoe".

Example 2

Input: s = "erase*"

Output: ""

The entire string is removed, so we return an empty string.

Constraints

1 <= s.length <= 105
s consists of lowercase English letters and stars *.
The operation above can be performed on s.

Solution Approach

Use a Stack for Efficient Removal

By iterating through the string and using a stack, we push characters into the stack until we encounter a star. Upon encountering a star, we pop the top element from the stack, simulating the removal of the most recent character.

Simulate Left-to-Right Traversal

The solution involves iterating through the string from left to right, performing stack operations for each character and star. This ensures that all removals are handled in the correct order, and we efficiently remove characters as required.

Final String Construction

Once all characters are processed, the stack will contain the remaining characters in order. We can then construct the result string by joining the stack's elements into a final string without stars.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution is O(n), where n is the length of the string, as each character is processed once. The space complexity is O(n), which accounts for storing the characters in the stack during the traversal.

What Interviewers Usually Probe

Look for the candidate's understanding of stack-based state management.
Check if they optimize space and time complexity using the stack pattern.
Ask them to explain the stack's role in the problem-solving process.

Common Pitfalls or Variants

Common pitfalls

Forgetting to pop the stack when encountering a star, leading to incorrect results.
Misunderstanding the traversal order and removing characters from the wrong position.
Assuming the string might be empty after removals, which could lead to incorrect handling.

Follow-up variants

Modify the problem to handle uppercase letters as well.
Introduce multiple stars consecutively, testing how the solution handles continuous removals.
Change the approach to use an array instead of a stack, observing the trade-offs in space and time.

FAQ

How do I remove characters based on stars in this problem?

You can use a stack to handle the characters and pop them when a star appears, simulating the required removals efficiently.

What data structure should I use for efficient star removal?

A stack is ideal for this problem because it allows for quick removal of the most recently added character, which matches the problem's removal process.

How does the stack help solve this problem?

The stack allows you to manage characters as you traverse the string. When a star is encountered, the most recent character is popped, which efficiently handles the removal.

What is the time complexity of the stack-based approach?

The time complexity is O(n), where n is the length of the string, as each character is processed once during the traversal.

What are some common mistakes in this problem?

A common mistake is failing to pop the stack when a star is encountered, or removing characters in the wrong order due to improper traversal.

terminal

Solution

Solution 1: Stack Simulation

We can use a stack to simulate the operation process. Traverse the string $s$, and if the current character is not an asterisk, push it onto the stack; if the current character is an asterisk, pop the top element from the stack.

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class Solution:
    def removeStars(self, s: str) -> str:
        ans = []
        for c in s:
            if c == '*':
                ans.pop()
            else:
                ans.append(c)
        return ''.join(ans)

Continue Practicing

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