How do you solve Resulting String After Adjacent Removals on LeetCode?

Resulting String After Adjacent Removals (LeetCode #3561) 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: Adjacent pairs can be removed in constant time with a stack.

What is the time complexity of Resulting String After Adjacent Removals?

The optimal solution for Resulting String After Adjacent Removals runs in O(n) time and uses O(n) space. The stack allows for efficient management of removals, processing each character in a single pass.

What is the brute force solution for Resulting String After Adjacent Removals?

The brute force approach for Resulting String After Adjacent Removals is Brute Force, with O(n²) time complexity and O(1) space complexity. Repeatedly scan the string for adjacent pairs and remove them until no more can be found. This method is straightforward but inefficient as it may require multiple passes through the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Resulting String After Adjacent Removals?

Resulting String After Adjacent Removals uses the following concepts: String, Stack, Simulation. The recommended patterns to study are: Stack, Greedy.

What are the interview tips for Resulting String After Adjacent Removals?

Explain your thought process clearly. Consider edge cases and constraints. Practice similar problems to build intuition.

What are common mistakes when solving Resulting String After Adjacent Removals?

Not considering the circular nature of the alphabet. Failing to handle edge cases like empty strings.

#3561

Resulting String After Adjacent Removals

Medium

StringStackSimulationStackGreedy

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)

Use a stack to efficiently manage removals. Push characters onto the stack and check the top for adjacent pairs, allowing for quick removals.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty stack.
2Step 2: Traverse each character in the string; push it onto the stack if it doesn't form an adjacent pair with the top of the stack.
3Step 3: If it does form a pair, pop the stack (remove the top character).

solution.py8 lines

1def removeAdjacent(s):
2    stack = []
3    for char in s:
4        if stack and (abs(ord(stack[-1]) - ord(char)) == 1 or (stack[-1], char) in [('a', 'z'), ('z', 'a')]):
5            stack.pop()
6        else:
7            stack.append(char)
8    return ''.join(stack)

ℹ

Complexity note: The stack allows for efficient management of removals, processing each character in a single pass.

1Adjacent pairs can be removed in constant time with a stack.
2Using a stack helps manage the order of characters efficiently.

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