How do you solve Remove All Adjacent Duplicates In String on LeetCode?

Remove All Adjacent Duplicates In String (LeetCode #1047) 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 duplicates efficiently.

What is the brute force solution for Remove All Adjacent Duplicates In String?

The brute force approach for Remove All Adjacent Duplicates In String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves repeatedly scanning the string to find and remove adjacent duplicates until no more can be 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 Adjacent Duplicates In String?

Remove All Adjacent Duplicates In String uses the following concepts: String, Stack. The recommended patterns to study are: Stack, Greedy Algorithm.

What are the interview tips for Remove All Adjacent Duplicates In String?

Explain your thought process clearly while coding. Consider edge cases and test your code with different inputs. Be prepared to discuss the time and space complexity of your solution.

What are common mistakes when solving Remove All Adjacent Duplicates In String?

Not using a stack and trying to manipulate the string directly. Overlooking edge cases where multiple duplicates may be adjacent.

#1047

Remove All Adjacent Duplicates In String

Easy

StringStackStackGreedy Algorithm

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 allows us to efficiently manage the characters as we process the string. When we encounter a character, we check if it matches the top of the stack. If it does, we pop the stack (remove the duplicate); if not, we push the character onto the stack. This way, we only traverse the string once.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty stack.
2Step 2: Loop through each character in the string.
3Step 3: If the stack is not empty and the top of the stack equals the current character, pop the stack.
4Step 4: If they are not equal, push the current character onto the stack.
5Step 5: After processing all characters, join the stack to form the final string.

solution.py8 lines

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

ℹ

Complexity note: The time complexity is linear because we traverse the string once, and the space complexity is also linear due to the stack potentially holding all characters in the worst case.

1Using a stack helps manage duplicates efficiently.
2The problem can be solved in a single pass with a stack.

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