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

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Find the Longest Valid Obstacle Course at Each Position?

This problem uses Array, Binary Search, Binary Indexed Tree. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Find the Longest Valid Obstacle Course at Each Position?

Insight 1: The problem is essentially about finding the length of the Longest Non-decreasing Subsequence (LNDS) ending at each index, with a constraint to include the obstacle at that index. Insight 2: We can use binary search to efficiently maintain and update the LNDS. Instead of storing the entire LNDS sequence, we only need to store the smallest tail element for each possible subsequence length.

Which companies ask LeetCode 1964. Find the Longest Valid Obstacle Course at Each Position in interviews?

This problem is frequently asked at Morgan Stanley.

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Find the Longest Valid Obstacle Course at Each Position - LeetCode 1964 Solution

Find the Longest Valid Obstacle Course at Each Position - Complete Solution Guide

Find the Longest Valid Obstacle Course at Each Position is LeetCode problem 1964, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You want to build some obstacle courses. You are given a 0-indexed integer array obstacles of length n , where obstacles[i] describes the height of the i th obstacle. For every index i between 0 and n - 1 ( inclusive ), find the length of the longest obstacle course in obstacles such that: You choose any number of obstacles between 0 and i inclusive . You must include the i th obstacle in the course. You must put the chosen obstacles in the same order as they appear in obstacles . Every obstacle

Detailed Explanation

The problem asks us to find, for each position `i` in an array `obstacles`, the length of the longest obstacle course ending at that position, subject to the following constraints: the course must include the obstacle at index `i`, obstacles must be chosen in their original order, and the height of each obstacle (except the first) must be greater than or equal to the height of the previous one. In simpler terms, we are looking for the length of the Longest Non-decreasing Subsequence (LNDS) ending at each index `i`, where the element at index `i` is definitely included in the LNDS.

Solution Approach

The provided solution uses a dynamic programming approach combined with binary search. It iterates through the `obstacles` array, maintaining a list `lis` that represents the smallest tail elements of all possible non-decreasing subsequences encountered so far. For each obstacle at index `i`, the solution uses binary search (`bisect_right` in Python, `binarySearchRight` in Java, `upper_bound` in C++, `bisect_right` in C) to find the position where the current obstacle can either extend an existing subsequence or replace a larger tail element to form a smaller tail element for a subsequence of the same length. The index returned by the binary search plus one gives the length of the longest obstacle course ending at index `i`.

Step-by-Step Algorithm

Step 1: Initialize an empty list `lis` to store the smallest tail elements of non-decreasing subsequences.
Step 2: Initialize an empty list `ans` to store the length of the longest obstacle course at each position.
Step 3: Iterate through the `obstacles` array from left to right.
Step 4: For each obstacle, use binary search to find the index where it can be inserted into the `lis` list while maintaining the non-decreasing property and smallest tail elements.
Step 5: If the binary search returns an index equal to the length of `lis`, it means the current obstacle is larger than all existing tail elements and extends the longest subsequence found so far. Append it to `lis`.
Step 6: If the binary search returns an index within the bounds of `lis`, it means the current obstacle can replace a larger tail element to form a non-decreasing subsequence of the same length but with a smaller tail element. Replace the element at that index with the current obstacle.
Step 7: Append the index returned by the binary search plus one to the `ans` list, representing the length of the longest obstacle course ending at the current position.
Step 8: After iterating through all obstacles, return the `ans` list.

Key Insights

Insight 1: The problem is essentially about finding the length of the Longest Non-decreasing Subsequence (LNDS) ending at each index, with a constraint to include the obstacle at that index.
Insight 2: We can use binary search to efficiently maintain and update the LNDS. Instead of storing the entire LNDS sequence, we only need to store the smallest tail element for each possible subsequence length.
Insight 3: The `lis` list maintains the increasing order of obstacle heights and can be updated efficiently using binary search to find the correct position to insert or replace elements.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search, Binary Indexed Tree.

Companies

Asked at: Morgan Stanley.

Find the Longest Valid Obstacle Course at Each Position | LeetCode 1964

Hard

Array Binary Search Binary Indexed Tree

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public int[] longestObstacleCourseAtEachPosition(int[] obstacles) {
        List<Integer> lis = new ArrayList<>();
        int[] ans = new int[obstacles.length];
        for (int i = 0; i < obstacles.length; i++) {
            int obstacle = obstacles[i];
            int idx = binarySearchRight(lis, obstacle);
            if (idx == lis.size()) {
                lis.add(obstacle);
            } else {
                lis.set(idx, obstacle);
            }
            ans[i] = idx + 1;
        }
        return ans;
    }

    private int binarySearchRight(List<Integer> list, int target) {
        int left = 0;
        int right = list.size();
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (list.get(mid) <= target) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return left;
    }
}

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Problem Info

Description

Problem 1964: Find the Longest Valid Obstacle Course at Each Position - Detailed explanation and solution approaches.

Topics

Array Binary Search Binary Indexed Tree

Companies

Morgan Stanley

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1964

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize an empty list `lis` to store the smallest tail elements of non-decreasing subsequences.
2
Step 2: Initialize an empty list `ans` to store the length of the longest obstacle course at each position.
3
Step 3: Iterate through the `obstacles` array from left to right.
4
Step 4: For each obstacle, use binary search to find the index where it can be inserted into the `lis` list while maintaining the non-decreasing property and smallest tail elements.
5
Step 5: If the binary search returns an index equal to the length of `lis`, it means the current obstacle is larger than all existing tail elements and extends the longest subsequence found so far. Append it to `lis`.
6
Step 6: If the binary search returns an index within the bounds of `lis`, it means the current obstacle can replace a larger tail element to form a non-decreasing subsequence of the same length but with a smaller tail element. Replace the element at that index with the current obstacle.
7
Step 7: Append the index returned by the binary search plus one to the `ans` list, representing the length of the longest obstacle course ending at the current position.
8
Step 8: After iterating through all obstacles, return the `ans` list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operations are the loop through the input array (O(n)) and the binary search within the loop (O(log n)). Therefore, the overall time complexity is O(n log n). The `lis` array stores the smallest tail elements which, in the worst-case scenario where all obstacle values are unique and increasing, could contain all elements of the original array, thus taking up O(n) space. The `ans` array also takes O(n) space.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming with O(n^2) time complexity

A basic dynamic programming approach involves creating a `dp` array where `dp[i]` stores the length of the longest non-decreasing subsequence ending at index `i`. For each `i`, iterate from `0` to `i-1` and check if `obstacles[j] <= obstacles[i]`. If so, update `dp[i] = max(dp[i], dp[j] + 1)`. The answer would be `dp[i]` after processing each index `i`.

Time: O(n^2)Space: O(n)

Trade-offs: This approach is simpler to understand but less efficient than the binary search approach. It's suitable for smaller input sizes but becomes inefficient for larger arrays due to its quadratic time complexity. The advantage is easier implementation.

✨

Key Insights

1Insight 1: The problem is essentially about finding the length of the Longest Non-decreasing Subsequence (LNDS) ending at each index, with a constraint to include the obstacle at that index.
2Insight 2: We can use binary search to efficiently maintain and update the LNDS. Instead of storing the entire LNDS sequence, we only need to store the smallest tail element for each possible subsequence length.
3Insight 3: The `lis` list maintains the increasing order of obstacle heights and can be updated efficiently using binary search to find the correct position to insert or replace elements.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: Empty input array. The solution implicitly handles this case by returning an empty array as the loop will not execute.
Edge case 2: Obstacles are in strictly increasing order. The `lis` array will simply contain all the obstacles, and the `ans` array will contain increasing values from 1 to `n`.
Edge case 3: Obstacles are in strictly decreasing order. The `lis` array will always have a length of 1, and elements are always replaced with smaller ones. The `ans` array will contain all 1s.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect binary search implementation. Using `bisect_left` instead of `bisect_right` (or the equivalent in other languages) will result in an incorrect LNDS length.

✗Common mistake 2: Not including the current obstacle in the subsequence. The algorithm must ensure that the obstacle at the current index is part of the subsequence being constructed.

✗Common mistake 3: Confusing longest increasing subsequence with longest non-decreasing subsequence. The obstacles can be of the same height, so equality must be considered in the comparison.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Stock price analysis. Identifying the longest period where the stock price is consistently increasing or staying the same can be useful for investment strategies.

🏦Real-world scenario 2: Task scheduling. Finding the longest sequence of tasks that can be executed in order, given their dependencies and resource requirements, where the dependencies can be represented as increasing values.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 12, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Find the Longest Valid Obstacle Course at Each Position"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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