What is the time complexity of Delivering Boxes from Storage to Ports?

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

What data structures are used to solve Delivering Boxes from Storage to Ports?

This problem uses Array, Dynamic Programming, Segment Tree, Queue, Heap (Priority Queue), Prefix Sum, Monotonic Queue. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Delivering Boxes from Storage to Ports?

Insight 1: Dynamic programming is suitable because the optimal solution for delivering the first 'i' boxes depends on the optimal solutions for delivering the first 'j' boxes (where j < i). Insight 2: Monotonic queue optimization is crucial to reduce the time complexity of the dynamic programming approach from O(n^2) to O(n), by efficiently maintaining the minimum cost within a sliding window of valid starting points for a trip.

Which companies ask LeetCode 1687. Delivering Boxes from Storage to Ports in interviews?

This problem is frequently asked at Nutanix.

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Delivering Boxes from Storage to Ports - LeetCode 1687 Solution

Q: How to solve Delivering Boxes from Storage to Ports?

The solution uses dynamic programming with a monotonic queue for optimization. `dp[i]` stores the minimum number of trips to deliver the first 'i' boxes. The algorithm iterates through each box, considering it as the end of a possible trip. For each box 'i', it finds the best starting box 'j' for the last trip. The cost of a trip from 'j' to 'i' involves a round trip (2 trips) and the number of port changes encountered while delivering boxes 'j' to 'i'. The monotonic queue efficiently finds the optimal 'j' to minimize the total number of trips. We use prefix sums of weights and port changes to quickly calculate weights and port change counts for given ranges.

Q: Which companies ask LeetCode 1687. Delivering Boxes from Storage to Ports in interviews?

This problem is frequently asked at Nutanix.

Delivering Boxes from Storage to Ports - Complete Solution Guide

Delivering Boxes from Storage to Ports is LeetCode problem 1687, 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 have the task of delivering some boxes from storage to their ports using only one ship. However, this ship has a limit on the number of boxes and the total weight that it can carry. You are given an array boxes , where boxes[i] = [ports i , weight i ] , and three integers portsCount , maxBoxes , and maxWeight . ports i is the port where you need to deliver the i th box and weights i is the weight of the i th box. portsCount is the number of ports. maxBoxes and maxWeight are the respecti

Detailed Explanation

The problem asks us to find the minimum number of trips a ship needs to make to deliver boxes from a storage to various ports. The boxes must be delivered in the given order. The ship has a limited capacity in terms of the number of boxes and the total weight it can carry. Each trip involves loading some boxes, visiting the necessary ports in order, and returning to the storage. The objective is to minimize the total number of these trips.

Solution Approach

The solution uses dynamic programming with a monotonic queue for optimization. `dp[i]` stores the minimum number of trips to deliver the first 'i' boxes. The algorithm iterates through each box, considering it as the end of a possible trip. For each box 'i', it finds the best starting box 'j' for the last trip. The cost of a trip from 'j' to 'i' involves a round trip (2 trips) and the number of port changes encountered while delivering boxes 'j' to 'i'. The monotonic queue efficiently finds the optimal 'j' to minimize the total number of trips. We use prefix sums of weights and port changes to quickly calculate weights and port change counts for given ranges.

Step-by-Step Algorithm

Step 1: Calculate `prefix_weights`: This array stores the cumulative weights of the boxes up to each index. `prefix_weights[i]` represents the sum of the weights of boxes from index 0 to i-1. This allows us to quickly calculate the total weight of any subarray of boxes.
Step 2: Calculate `diffs`: This array stores the cumulative number of port changes up to each index. `diffs[i]` represents the total number of port changes when delivering boxes from 0 to i-1. This helps to quickly determine the cost of trips between different sets of boxes.
Step 3: Initialize `dp`: `dp[i]` will store the minimum number of trips needed to deliver the first 'i' boxes. `dp[0]` is initialized to 0 (no trips needed for zero boxes).
Step 4: Initialize monotonic queue `dq`: The monotonic queue will store the indices of boxes that are potential starting points for the last trip. It maintains the property that `dp[j] - diffs[j+1]` is monotonically increasing within the queue, helping to efficiently find the minimum cost.
Step 5: Iterate through each box from 1 to n: For each box 'i', we aim to find the optimal 'j' such that `dp[i] = min(dp[j] + 2 + diffs[i] - diffs[j+1])` for all valid 'j'.
Step 6: Maintain the monotonic queue: Remove elements from the front of the queue if they are no longer valid (i.e., the number of boxes or total weight between 'j' and 'i' exceeds the limits). Remove elements from the back of the queue if the current 'dp[i] - diffs[i+1]' is less than or equal to the value at the back of the queue (maintaining the monotonically increasing property).
Step 7: Find the optimal 'j': The optimal 'j' is the element at the front of the queue after the invalid elements have been removed.
Step 8: Update `dp[i]`: Calculate `dp[i]` using the optimal 'j' and the cost of the last trip (2 + `diffs[i]` - `diffs[j+1]`).
Step 9: Add 'i' to the queue: If i < n, add the current box 'i' index into the queue. This guarantees that all potential starting points for subsequent calculations are available in the queue for later calculations
Step 10: Return `dp[n]`: The value of `dp[n]` represents the minimum number of trips needed to deliver all 'n' boxes.

Key Insights

Insight 1: Dynamic programming is suitable because the optimal solution for delivering the first 'i' boxes depends on the optimal solutions for delivering the first 'j' boxes (where j < i).
Insight 2: Monotonic queue optimization is crucial to reduce the time complexity of the dynamic programming approach from O(n^2) to O(n), by efficiently maintaining the minimum cost within a sliding window of valid starting points for a trip.
Insight 3: Pre-calculating prefix sums of weights and the number of port changes allows us to compute the weight and cost of trips in O(1) time, which are vital for efficiency.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Segment Tree, Queue, Heap (Priority Queue), Prefix Sum, Monotonic Queue.

Companies

Asked at: Nutanix.

Delivering Boxes from Storage to Ports | LeetCode 1687

Hard

Array Dynamic Programming Segment Tree Queue Heap (Priority Queue)Prefix Sum Monotonic Queue

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

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.Deque;

class Solution {
    public int boxDelivering(int[][] boxes, int portsCount, int maxBoxes, int maxWeight) {
        int n = boxes.length;

        int[] prefix_weights = new int[n + 1];
        for (int i = 0; i < n; i++) {
            prefix_weights[i + 1] = prefix_weights[i] + boxes[i][1];
        }

        int[] diffs = new int[n + 1];
        for (int i = 1; i < n; i++) {
            if (boxes[i][0] != boxes[i - 1][0]) {
                diffs[i + 1] = diffs[i] + 1;
            } else {
                diffs[i + 1] = diffs[i];
            }
        }

        int[] dp = new int[n + 1];

        Deque<Integer> dq = new ArrayDeque<>();
        dq.add(0);

        for (int i = 1; i < n + 1; i++) {
            while (!dq.isEmpty() && (i - dq.peekFirst() > maxBoxes || prefix_weights[i] - prefix_weights[dq.peekFirst()] > maxWeight)) {
                dq.pollFirst();
            }

            int best_j = dq.peekFirst();

            int min_g = dp[best_j] - diffs[best_j + 1];
            dp[i] = 2 + diffs[i] + min_g;

            if (i < n) {
                int g_i = dp[i] - diffs[i + 1];
                while (!dq.isEmpty() && (dp[dq.peekLast()] - diffs[dq.peekLast() + 1]) >= g_i) {
                    dq.pollLast();
                }
                dq.addLast(i);
            }
        }

        return dp[n];
    }
}

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Problem Info

Description

Problem 1687: Delivering Boxes from Storage to Ports - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Segment Tree Queue Heap (Priority Queue)Prefix Sum Monotonic Queue

Companies

Nutanix

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1687

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

10 steps

1
Step 1: Calculate `prefix_weights`: This array stores the cumulative weights of the boxes up to each index. `prefix_weights[i]` represents the sum of the weights of boxes from index 0 to i-1. This allows us to quickly calculate the total weight of any subarray of boxes.
2
Step 2: Calculate `diffs`: This array stores the cumulative number of port changes up to each index. `diffs[i]` represents the total number of port changes when delivering boxes from 0 to i-1. This helps to quickly determine the cost of trips between different sets of boxes.
3
Step 3: Initialize `dp`: `dp[i]` will store the minimum number of trips needed to deliver the first 'i' boxes. `dp[0]` is initialized to 0 (no trips needed for zero boxes).
4
Step 4: Initialize monotonic queue `dq`: The monotonic queue will store the indices of boxes that are potential starting points for the last trip. It maintains the property that `dp[j] - diffs[j+1]` is monotonically increasing within the queue, helping to efficiently find the minimum cost.
5
Step 5: Iterate through each box from 1 to n: For each box 'i', we aim to find the optimal 'j' such that `dp[i] = min(dp[j] + 2 + diffs[i] - diffs[j+1])` for all valid 'j'.
6
Step 6: Maintain the monotonic queue: Remove elements from the front of the queue if they are no longer valid (i.e., the number of boxes or total weight between 'j' and 'i' exceeds the limits). Remove elements from the back of the queue if the current 'dp[i] - diffs[i+1]' is less than or equal to the value at the back of the queue (maintaining the monotonically increasing property).
7
Step 7: Find the optimal 'j': The optimal 'j' is the element at the front of the queue after the invalid elements have been removed.
8
Step 8: Update `dp[i]`: Calculate `dp[i]` using the optimal 'j' and the cost of the last trip (2 + `diffs[i]` - `diffs[j+1]`).
9
Step 9: Add 'i' to the queue: If i < n, add the current box 'i' index into the queue. This guarantees that all potential starting points for subsequent calculations are available in the queue for later calculations
10
Step 10: Return `dp[n]`: The value of `dp[n]` represents the minimum number of trips needed to deliver all 'n' boxes.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The algorithm iterates through each box once (O(n)). The monotonic queue operations (adding and removing elements) take O(1) amortized time per box, as each box is added and removed at most once. The prefix sum and diff array calculations are also O(n). Therefore, the overall time complexity is O(n). The `prefix_weights`, `diffs`, `dp` arrays and the monotonic queue each take O(n) space, resulting in an overall space complexity of O(n).

📖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 without Monotonic Queue

A simpler dynamic programming approach without the monotonic queue optimization. It iterates through all possible starting points 'j' for each ending point 'i' and calculates the cost directly.

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

Trade-offs: This approach is easier to understand and implement, but it's less efficient (O(n^2)) and will likely result in a Time Limit Exceeded error for larger input sizes. Use the monotonic queue for optimal performance.

✨

Key Insights

1Insight 1: Dynamic programming is suitable because the optimal solution for delivering the first 'i' boxes depends on the optimal solutions for delivering the first 'j' boxes (where j < i).
2Insight 2: Monotonic queue optimization is crucial to reduce the time complexity of the dynamic programming approach from O(n^2) to O(n), by efficiently maintaining the minimum cost within a sliding window of valid starting points for a trip.
3Insight 3: Pre-calculating prefix sums of weights and the number of port changes allows us to compute the weight and cost of trips in O(1) time, which are vital for efficiency.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Queue

FIFO structure - First In, First Out. Like a line at a store!

Edge Cases

Edge case 1: `boxes` is empty. The code implicitly handles this by returning 0 since the loops won't execute.
Edge case 2: `maxBoxes` or `maxWeight` is very small. The monotonic queue ensures that only valid starting points are considered, so even if the limits are restrictive, the algorithm will still find the optimal solution.
Edge case 3: A single box exceeds `maxWeight`. The problem statement says `1 <= weights_i <= maxWeight`, so this will not occur. The code will break if any weight exceeds maxWeight, so an additional validation would be needed in that case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the number of port changes. It's crucial to use the pre-calculated `diffs` array to ensure the correct count is used in the DP recurrence.

✗Common mistake 2: Failing to maintain the monotonically increasing property of the `dp[j] - diffs[j+1]` values in the queue. This is essential for the queue to correctly identify the minimum cost starting point.

✗Common mistake 3: Forgetting to remove invalid elements from the front of the queue. Elements become invalid when the box count or weight exceeds the `maxBoxes` or `maxWeight` limit

✅

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: Optimizing delivery routes for a logistics company with capacity constraints. Each delivery location is a 'port', and each package is a 'box' with a certain 'weight'. The goal is to minimize the number of trips while adhering to vehicle capacity.

🏦Real-world scenario 2: Resource allocation in a cloud computing environment. Tasks (boxes) need to be assigned to servers (ports) with limited resources (maxBoxes, maxWeight). The objective is to minimize the number of server restarts (trips) required to complete all tasks in the specified order.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: August 7, 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 13

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

@CodeNinjaJul 6

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

@TechLeadJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Delivering Boxes from Storage to Ports"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?