What is the time complexity of Time to Cross a Bridge?

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

What data structures are used to solve Time to Cross a Bridge?

This problem uses Array, Heap (Priority Queue), Simulation. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Time to Cross a Bridge?

Insight 1: The core of the problem is managing the order of workers crossing the bridge based on their 'inefficiency' and availability on either side of the bridge. This requires maintaining multiple queues to track workers waiting on each side. Insight 2: Prioritization is key. The problem specifies that the *least efficient* worker should be prioritized. A max-heap (priority queue) is suitable for this because it easily provides access to the element with the largest 'inefficiency' value.

Which companies ask LeetCode 2532. Time to Cross a Bridge in interviews?

This problem is frequently asked at LinkedIn.

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Time to Cross a Bridge - LeetCode 2532 Solution

Time to Cross a Bridge - Complete Solution Guide

Time to Cross a Bridge is LeetCode problem 2532, 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

There are k workers who want to move n boxes from the right (old) warehouse to the left (new) warehouse. You are given the two integers n and k , and a 2D integer array time of size k x 4 where time[i] = [right i , pick i , left i , put i ] . The warehouses are separated by a river and connected by a bridge. Initially, all k workers are waiting on the left side of the bridge. To move the boxes, the i th worker can do the following: Cross the bridge to the right side in right i minutes. Pick a bo

Detailed Explanation

The problem simulates a scenario where 'k' workers transport 'n' boxes from a right warehouse to a left warehouse across a bridge. Each worker has specific times for crossing the bridge to the right, picking up a box, crossing to the left, and putting down the box. Only one worker can cross the bridge at a time. Workers are prioritized based on their efficiency (or inefficiency, more accurately) – the less efficient worker crossing first, prioritizing the side that has completed the pick-up phase, before considering a worker on the left side. Efficiency is determined by the sum of their right and left crossing times; if these sums are equal, the worker with the higher index is considered less efficient. The goal is to find the time when the last box reaches the left side of the bridge.

Solution Approach

The solution employs a simulation approach. It uses four priority queues (heaps): `wait_left`, `wait_right`, `picking_right`, and `putting_left`. `wait_left` and `wait_right` store workers waiting on the left and right sides of the bridge, respectively, prioritized by inefficiency (largest left+right first). `picking_right` and `putting_left` store workers who are currently picking up boxes on the right and putting down boxes on the left, respectively, prioritized by completion time (smallest completion time first). The algorithm iterates until all boxes have crossed to the left side. In each iteration, it checks if any workers have finished picking up or putting down boxes and moves them to the appropriate waiting queues. Then, it prioritizes workers from `wait_right` to cross the bridge if available, otherwise from `wait_left` (if there are still boxes to dispatch). If both waiting queues are empty, it advances the time to the next event in either `picking_right` or `putting_left`.

Step-by-Step Algorithm

Step 1: Initialize four priority queues: `wait_left`, `wait_right`, `picking_right`, and `putting_left`. Populate `wait_left` with the initial 'k' workers and their corresponding inefficiencies. In Python, the `wait_left` initialization utilizes the negative of the `left + right` values for proper max-heap behavior.
Step 2: Initialize `current_time` to 0, `boxes_dispatched` to 0, and `boxes_crossed_to_left` to 0.
Step 3: While `boxes_crossed_to_left` < 'n', repeat the following steps:
Step 4: Move workers from `picking_right` to `wait_right` if their pickup completion time is less than or equal to `current_time`.
Step 5: Move workers from `putting_left` to `wait_left` if their putdown completion time is less than or equal to `current_time`.
Step 6: If `wait_right` is not empty, select the most inefficient worker (highest priority in max heap), move them to `putting_left` after crossing the bridge to the left, increment `boxes_crossed_to_left`, and update `current_time`.
Step 7: Else, if `wait_left` is not empty and `boxes_dispatched` < 'n', select the most inefficient worker, move them to `picking_right` after crossing the bridge to the right, increment `boxes_dispatched`, and update `current_time`.
Step 8: If both `wait_right` and `wait_left` are empty, advance `current_time` to the time of the next event (either a worker finishes picking up or putting down) if there are pending events. Break if there are no pending events.
Step 9: Return `current_time`.

Key Insights

Insight 1: The core of the problem is managing the order of workers crossing the bridge based on their 'inefficiency' and availability on either side of the bridge. This requires maintaining multiple queues to track workers waiting on each side.
Insight 2: Prioritization is key. The problem specifies that the *least efficient* worker should be prioritized. A max-heap (priority queue) is suitable for this because it easily provides access to the element with the largest 'inefficiency' value.
Insight 3: The simulation requires tracking the completion times of workers who are currently picking up or putting down boxes. We use min-heaps for these tasks to efficiently determine when workers become available.
Insight 4: It's essential to consider edge cases where one side is empty, and the other side has workers available, or where all boxes have been dispatched from one side. We need to avoid sending unnecessary workers across the bridge if all the boxes are already on their way.
Insight 5: The question is asking the time when the LAST box reaches the left side, not when the last worker completes their action. Therefore, we should stop the simulation when all boxes reach the left side of the bridge.

Complexity Analysis

Time Complexity: O(n log k)

Space Complexity: O(k)

Topics

This problem involves: Array, Heap (Priority Queue), Simulation.

Companies

Asked at: LinkedIn.

Time to Cross a Bridge | LeetCode 2532

Hard

Array Heap (Priority Queue)Simulation

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

Solution Code

Available in:

import java.util.PriorityQueue;

class Solution {
    public int findCrossingTime(int n, int k, int[][] time) {
        PriorityQueue<int[]> wait_left = new PriorityQueue<>((a, b) -> {
            if (a[0] != b[0]) {
                return a[0] - b[0];
            } else {
                return a[1] - b[1];
            }
        });

        for (int i = 0; i < k; i++) {
            wait_left.offer(new int[]{- (time[i][0] + time[i][2]), -i});
        }

        PriorityQueue<int[]> wait_right = new PriorityQueue<>((a, b) -> {
            if (a[0] != b[0]) {
                return a[0] - b[0];
            } else {
                return a[1] - b[1];
            }
        });

        PriorityQueue<int[]> picking_right = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        PriorityQueue<int[]> putting_left = new PriorityQueue<>((a, b) -> a[0] - b[0]);

        int current_time = 0;
        int boxes_dispatched = 0;
        int boxes_crossed_to_left = 0;

        while (boxes_crossed_to_left < n) {
            while (!picking_right.isEmpty() && picking_right.peek()[0] <= current_time) {
                int[] worker = picking_right.poll();
                int worker_idx = worker[1];
                wait_right.offer(new int[]{- (time[worker_idx][0] + time[worker_idx][2]), -worker_idx});
            }

            while (!putting_left.isEmpty() && putting_left.peek()[0] <= current_time) {
                int[] worker = putting_left.poll();
                int worker_idx = worker[1];
                wait_left.offer(new int[]{- (time[worker_idx][0] + time[worker_idx][2]), -worker_idx});
            }

            if (!wait_right.isEmpty()) {
                int[] worker_neg = wait_right.poll();
                int worker_idx = -worker_neg[1];

                current_time += time[worker_idx][2];
                boxes_crossed_to_left += 1;

                putting_left.offer(new int[]{current_time + time[worker_idx][3], worker_idx});

            } else if (!wait_left.isEmpty() && boxes_dispatched < n) {
                int[] worker_neg = wait_left.poll();
                int worker_idx = -worker_neg[1];

                boxes_dispatched += 1;
                current_time += time[worker_idx][0];

                picking_right.offer(new int[]{current_time + time[worker_idx][1], worker_idx});

            } else {
                int next_event_time = Integer.MAX_VALUE;
                if (!picking_right.isEmpty()) {
                    next_event_time = Math.min(next_event_time, picking_right.peek()[0]);
                }
                if (!putting_left.isEmpty()) {
                    next_event_time = Math.min(next_event_time, putting_left.peek()[0]);
                }

                if (next_event_time == Integer.MAX_VALUE) {
                    break;
                }

                current_time = next_event_time;
            }
        }

        return current_time;
    }
}

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Problem Info

Description

Problem 2532: Time to Cross a Bridge - Detailed explanation and solution approaches.

Topics

Array Heap (Priority Queue)Simulation

Companies

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2532

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

9 steps

1
Step 1: Initialize four priority queues: `wait_left`, `wait_right`, `picking_right`, and `putting_left`. Populate `wait_left` with the initial 'k' workers and their corresponding inefficiencies. In Python, the `wait_left` initialization utilizes the negative of the `left + right` values for proper max-heap behavior.
2
Step 2: Initialize `current_time` to 0, `boxes_dispatched` to 0, and `boxes_crossed_to_left` to 0.
3
Step 3: While `boxes_crossed_to_left` < 'n', repeat the following steps:
4
Step 4: Move workers from `picking_right` to `wait_right` if their pickup completion time is less than or equal to `current_time`.
5
Step 5: Move workers from `putting_left` to `wait_left` if their putdown completion time is less than or equal to `current_time`.
6
Step 6: If `wait_right` is not empty, select the most inefficient worker (highest priority in max heap), move them to `putting_left` after crossing the bridge to the left, increment `boxes_crossed_to_left`, and update `current_time`.
7
Step 7: Else, if `wait_left` is not empty and `boxes_dispatched` < 'n', select the most inefficient worker, move them to `picking_right` after crossing the bridge to the right, increment `boxes_dispatched`, and update `current_time`.
8
Step 8: If both `wait_right` and `wait_left` are empty, advance `current_time` to the time of the next event (either a worker finishes picking up or putting down) if there are pending events. Break if there are no pending events.
9
Step 9: Return `current_time`.

💡

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 k)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(k)

📝

Why?

Each worker can cross the bridge at most a constant number of times (proportional to n). The heap operations (push and pop) take O(log k) time, where k is the number of workers. Since each box results in one crossing of a box from right to left, the total number of operations is proportional to n, yielding O(n log k). The space complexity is dominated by the size of the priority queues, which can hold at most 'k' workers.

📖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

Simulation with List/Array

Instead of priority queues, you could use arrays or lists to represent the waiting and working workers. You'd then need to iterate through these lists to find the most inefficient worker or the worker who finishes the earliest.

Time: O(n*k)Space: O(k)

Trade-offs: This approach would be simpler to implement but less efficient because finding the minimum/maximum element in a list takes O(k) time, resulting in an overall time complexity of O(n*k).

✨

Key Insights

1Insight 1: The core of the problem is managing the order of workers crossing the bridge based on their 'inefficiency' and availability on either side of the bridge. This requires maintaining multiple queues to track workers waiting on each side.
2Insight 2: Prioritization is key. The problem specifies that the *least efficient* worker should be prioritized. A max-heap (priority queue) is suitable for this because it easily provides access to the element with the largest 'inefficiency' value.
3Insight 3: The simulation requires tracking the completion times of workers who are currently picking up or putting down boxes. We use min-heaps for these tasks to efficiently determine when workers become available.
4Insight 4: It's essential to consider edge cases where one side is empty, and the other side has workers available, or where all boxes have been dispatched from one side. We need to avoid sending unnecessary workers across the bridge if all the boxes are already on their way.
5Insight 5: The question is asking the time when the LAST box reaches the left side, not when the last worker completes their action. Therefore, we should stop the simulation when all boxes reach the left side of the bridge.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: `n = 1`, `k = 1`. The solution correctly simulates the single worker's journey, ensuring the correct time is returned.
Edge case 2: `n` is much larger than `k`. The solution handles the case where all workers are continually busy transporting boxes, correctly managing the queues and time progression.
Edge case 3: All workers have the same `left + right` time. The secondary tie-breaker (worker index) in the priority queue comparator ensures the correct order based on the worker's index.
Edge case 4: One of the wait queues is empty for a long time. The algorithm correctly advances time to the next available worker.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly prioritizing workers. Failing to properly implement the inefficiency comparison (left + right time, and index as a tie-breaker) can lead to an incorrect simulation.

✗Common mistake 2: Neglecting to update the `current_time` when a worker finishes their task. This can cause the simulation to get out of sync.

✗Common mistake 3: Not considering the case where a worker has to wait for an event on the bridge. When all the wait queues are empty, advancing the time to the next event is critical.

✗Common mistake 4: Not considering when to stop the workers. The problem is about the last BOX reaches the left, not when the last worker puts down the box. Stopping when all boxes reach the left side is crucial for avoiding extra computations.

✅

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: Task scheduling in a multi-processor system. Workers represent processors, tasks represent boxes, and the bridge represents a shared resource. The algorithm efficiently schedules tasks to processors based on their processing speeds and resource availability.

🏦Real-world scenario 2: Optimizing traffic flow in a transportation network. Workers represent vehicles, boxes represent goods, and the bridge represents a bottleneck. The algorithm prioritizes vehicles to minimize congestion and delivery time.

🎮Real-world scenario 3: Assembly line process with limited work stations, prioritizing less efficient assembly stations to minimize overall process time.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: June 13, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

❓

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 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 "Time to Cross a Bridge"?

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 heap (priority queue) techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?