How do you solve Removing Stars From a String on LeetCode?

Removing Stars From a String (LeetCode #2390) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a stack helps manage the removal of characters efficiently.

What algorithm or data structure is used to solve Removing Stars From a String?

Removing Stars From a String uses the following concepts: String, Stack, Simulation. The recommended patterns to study are: Stack, Simulation.

What are the interview tips for Removing Stars From a String?

Explain your thought process clearly while coding. Consider edge cases and discuss them with the interviewer. Practice stack-related problems to become comfortable with their operations.

What are common mistakes when solving Removing Stars From a String?

Not considering edge cases where multiple stars are adjacent. Failing to manage the stack correctly, leading to incorrect results.

Removing Stars From a String — LeetCode #2390 (Medium)

Tags: String, Stack, Simulation

Related patterns: Stack, Simulation

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves iterating through the string and removing characters as we encounter stars. For each star, we find and remove the closest non-star character to its left, which can be inefficient.

The time complexity is O(n²) because for each star, we may need to traverse the result string to remove the last character, leading to a quadratic number of operations in the worst case.

Step 1: Initialize an empty result string.
Step 2: Iterate through each character in the string.
Step 3: If the character is not a star, add it to the result string.
Step 4: If the character is a star, remove the last character added to the result string (if any).

Input: 'leet**cod*e' 1. Initialize result = '' 2. Read 'l' -> result = 'l' 3. Read 'e' -> result = 'le' 4. Read 'e' -> result = 'lee' 5. Read 't' -> result = 'leet' 6. Read '*' -> result = 'lee' (remove 't') 7. Read '*' -> result = 'le' (remove 'e') 8. Read 'c' -> result = 'lec' 9. Read 'o' -> result = 'leco' 10. Read 'd' -> result = 'lecod' 11. Read '*' -> result = 'lecoe' (remove 'd') 12. Read 'e' -> result = 'lecoe'

Optimal Solution approach

Time complexity: O(n). Space complexity: O(n).

Using a stack allows us to efficiently manage the characters as we process the string. When we encounter a star, we simply pop the last character from the stack, which represents the closest non-star character to the left.

The time complexity is O(n) because we process each character in the string once. The space complexity is O(n) in the worst case when there are no stars, and all characters are pushed onto the stack.

Step 1: Initialize an empty stack.
Step 2: Iterate through each character in the string.
Step 3: If the character is not a star, push it onto the stack.
Step 4: If the character is a star, pop the top character from the stack (if the stack is not empty).
Step 5: After processing all characters, join the characters in the stack to form the result string.

Input: 'leet**cod*e' 1. Initialize stack = [] 2. Read 'l' -> stack = ['l'] 3. Read 'e' -> stack = ['l', 'e'] 4. Read 'e' -> stack = ['l', 'e', 'e'] 5. Read 't' -> stack = ['l', 'e', 'e', 't'] 6. Read '*' -> stack = ['l', 'e', 'e'] (pop 't') 7. Read '*' -> stack = ['l', 'e'] (pop 'e') 8. Read 'c' -> stack = ['l', 'e', 'c'] 9. Read 'o' -> stack = ['l', 'e', 'c', 'o'] 10. Read 'd' -> stack = ['l', 'e', 'c', 'o', 'd'] 11. Read '*' -> stack = ['l', 'e', 'c', 'o'] (pop 'd') 12. Final result = 'lecoe' from stack

Key Insights

Using a stack helps manage the removal of characters efficiently.
The problem can be visualized as a series of undo operations, which is what stacks are designed for.

Common Mistakes

Not considering edge cases where multiple stars are adjacent.
Failing to manage the stack correctly, leading to incorrect results.

Interview Tips

Explain your thought process clearly while coding.
Consider edge cases and discuss them with the interviewer.
Practice stack-related problems to become comfortable with their operations.

#2390

Removing Stars From a String

Medium

String↗Stack↗Simulation↗Stack↗Simulation↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty stack.
2Step 2: Iterate through each character in the string.
3Step 3: If the character is not a star, push it onto the stack.
4Step 4: If the character is a star, pop the top character from the stack (if the stack is not empty).
5Step 5: After processing all characters, join the characters in the stack to form the result string.

solution.py9 lines

1def removeStars(s):
2    stack = []
3    for char in s:
4        if char != '*':
5            stack.append(char)
6        else:
7            if stack:
8                stack.pop()
9    return ''.join(stack)

ℹ

Complexity note: The time complexity is O(n) because we process each character in the string once. The space complexity is O(n) in the worst case when there are no stars, and all characters are pushed onto the stack.

1Using a stack helps manage the removal of characters efficiently.
2The problem can be visualized as a series of undo operations, which is what stacks are designed for.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.