How do you solve Remove All Occurrences of a Substring on LeetCode?

Remove All Occurrences of a Substring (LeetCode #1910) 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: Removing a substring can create new opportunities for removal, so a single pass may not be sufficient.

What is the time complexity of Remove All Occurrences of a Substring?

The optimal solution for Remove All Occurrences of a Substring runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse the string once, and the space complexity is O(n) due to the storage of characters in the stack.

What is the brute force solution for Remove All Occurrences of a Substring?

The brute force approach for Remove All Occurrences of a Substring is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves repeatedly searching for the substring 'part' in 's' and removing it until no more occurrences are found. This is straightforward but inefficient as it may require multiple passes over the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Remove All Occurrences of a Substring?

Remove All Occurrences of a Substring uses the following concepts: String, Stack, Simulation. The recommended patterns to study are: Stack, String Manipulation.

What are the interview tips for Remove All Occurrences of a Substring?

Always think about edge cases, such as when the substring is at the start or end of the string. Consider using data structures like stacks for problems involving substrings or sequences. Explain your thought process clearly to the interviewer, especially when transitioning from brute force to optimal solutions.

What are common mistakes when solving Remove All Occurrences of a Substring?

Not considering that removing a substring can lead to new substrings being formed. Using inefficient string operations that lead to excessive time complexity.

#1910

Remove All Occurrences of a Substring

Medium

StringStackSimulationStackString Manipulation

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)

Using a stack-like approach allows us to build the result string character by character. This way, we can efficiently check if the last characters match the substring 'part' and remove them if they do, avoiding multiple passes over the string.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty list (acting as a stack) to build the result.
2Step 2: Iterate through each character in 's'.
3Step 3: Append each character to the stack.
4Step 4: Check if the last characters in the stack match 'part'. If they do, remove them from the stack.
5Step 5: Convert the stack back to a string and return it.

solution.py8 lines

1def removeOccurrences(s: str, part: str) -> str:
2    stack = []
3    part_length = len(part)
4    for char in s:
5        stack.append(char)
6        if ''.join(stack[-part_length:]) == part:
7            del stack[-part_length:]
8    return ''.join(stack)

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once, and the space complexity is O(n) due to the storage of characters in the stack.

1Removing a substring can create new opportunities for removal, so a single pass may not be sufficient.
2Using a stack allows us to efficiently manage the characters and check for the substring in a single traversal.

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