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

Remove All Adjacent Duplicates in String II (LeetCode #1209) 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 counts of characters efficiently.

What is the time complexity of Remove All Adjacent Duplicates in String II?

The optimal solution for Remove All Adjacent Duplicates in String II 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 stack storing characters and their counts.

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

The brute force approach for Remove All Adjacent Duplicates in String II 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 II?

Remove All Adjacent Duplicates in String II uses the following concepts: String, Stack. The recommended patterns to study are: Stack, Two Pointers.

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

Think about using data structures like stacks for problems involving sequences. Explain your thought process clearly while coding; it helps the interviewer understand your approach. Test your solution with edge cases, like strings with no duplicates or all characters being the same.

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

Not managing counts correctly when pushing to the stack. Forgetting to pop from the stack when the count reaches k.

#1209

Remove All Adjacent Duplicates in String II

Medium

StringStackStackTwo Pointers

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 and their counts. We can push characters onto the stack and keep track of their counts, removing them when they reach k.

⚙️

Algorithm

6 steps

1Step 1: Initialize an empty stack to hold pairs of characters and their counts.
2Step 2: Iterate through each character in the string.
3Step 3: If the stack is not empty and the top character matches the current character, increment the count.
4Step 4: If the count reaches k, pop the character from the stack.
5Step 5: If the character is new, push it onto the stack with a count of 1.
6Step 6: Construct the final string from the stack.

solution.py10 lines

1def removeDuplicates(s, k):
2    stack = []
3    for char in s:
4        if stack and stack[-1][0] == char:
5            stack[-1][1] += 1
6            if stack[-1][1] == k:
7                stack.pop()
8        else:
9            stack.append([char, 1])
10    return ''.join(char * count for char, count in 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 stack storing characters and their counts.

1Using a stack helps manage counts of characters efficiently.
2Adjacent duplicates can be removed 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.