What is the time complexity of Find the Longest Valid Obstacle Course at Each Position?

The optimal solution for Find the Longest Valid Obstacle Course at Each Position runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the binary search operation in the 'dp' list for each obstacle, while the space complexity is O(n) for storing the 'dp' list.

What is the brute force solution for Find the Longest Valid Obstacle Course at Each Position?

The brute force approach for Find the Longest Valid Obstacle Course at Each Position is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible subsequences of obstacles up to each index to find the longest valid obstacle course. This is straightforward but inefficient, as it examines every combination. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Find the Longest Valid Obstacle Course at Each Position?

Find the Longest Valid Obstacle Course at Each Position uses the following concepts: Array, Binary Search, Binary Indexed Tree. The recommended patterns to study are: Binary Search, Dynamic Programming.

What are the interview tips for Find the Longest Valid Obstacle Course at Each Position?

Always discuss the brute force solution first to show understanding. Explain your thought process clearly while transitioning to optimal solutions. Practice binary search problems to become comfortable with its application.

What are common mistakes when solving Find the Longest Valid Obstacle Course at Each Position?

Not considering the case where obstacles are equal in height. Failing to use binary search effectively, leading to O(n²) solutions.

#1964

Find the Longest Valid Obstacle Course at Each Position

Hard

ArrayBinary SearchBinary Indexed TreeBinary SearchDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

Using a binary search approach allows us to efficiently find the position of the current obstacle in a dynamic list of valid heights, leading to a faster solution. We maintain a list that tracks the minimum height for each valid length of the obstacle course.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty list 'dp' to store the minimum heights for each length of the obstacle course.
2Step 2: For each obstacle, use binary search to find the position in 'dp' where the current obstacle can replace or extend the sequence.
3Step 3: Update 'dp' accordingly and store the length of the valid course in 'ans'.

solution.py14 lines

1# Full working Python code
2import bisect
3
4def longestObstacleCourseAtEachPosition(obstacles):
5    dp = []
6    ans = []
7    for height in obstacles:
8        pos = bisect.bisect_right(dp, height)
9        if pos < len(dp):
10            dp[pos] = height
11        else:
12            dp.append(height)
13        ans.append(pos + 1)
14    return ans

ℹ

Complexity note: The time complexity is O(n log n) due to the binary search operation in the 'dp' list for each obstacle, while the space complexity is O(n) for storing the 'dp' list.

1The longest valid subsequence can be efficiently tracked using binary search.
2Maintaining a dynamic list of minimum heights allows for quick updates.

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