How do you solve Restore Finishing Order on LeetCode?

Restore Finishing Order (LeetCode #3668) 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 hash set allows for faster lookups.

What is the time complexity of Restore Finishing Order?

The optimal solution for Restore Finishing Order runs in O(n) time and uses O(n) space. Creating the hash set takes O(m) time, and filtering the order takes O(n), leading to linear complexity overall.

What is the brute force solution for Restore Finishing Order?

The brute force approach for Restore Finishing Order is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each friend against the finishing order one by one. This method is straightforward but inefficient as it involves nested loops. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Restore Finishing Order?

Restore Finishing Order uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Restore Finishing Order?

Explain your thought process clearly. Discuss time and space complexity. Consider edge cases and constraints.

What are common mistakes when solving Restore Finishing Order?

Not using a set for quick lookups. Assuming order will always be sorted.

#3668

Restore Finishing Order

Easy

ArrayHash TableHash MapArray

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 hash set for quick lookups of friends' IDs, allowing us to filter the order array efficiently.

⚙️

Algorithm

3 steps

1Step 1: Create a hash set from the friends array for O(1) lookups.
2Step 2: Iterate through the order array and check if each ID is in the hash set.
3Step 3: Collect matching IDs into the result array.

solution.py3 lines

1def restore_order(order, friends):
2    friend_set = set(friends)
3    return [friend for friend in order if friend in friend_set]

ℹ

Complexity note: Creating the hash set takes O(m) time, and filtering the order takes O(n), leading to linear complexity overall.

1Using a hash set allows for faster lookups.
2Filtering based on a set is more efficient than nested loops.

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