What is the time complexity of Capacity To Ship Packages Within D Days?

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

What data structures are used to solve Capacity To Ship Packages Within D Days?

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

What are the key insights for solving Capacity To Ship Packages Within D Days?

Insight 1: The problem can be solved using binary search because the possible capacities form a sorted search space. We can define a range of possible capacities (minimum and maximum possible capacity) and iteratively narrow down the search space. Insight 2: The 'can_ship' function is crucial. It checks whether it's possible to ship all packages within 'days' given a specific capacity. This is done by simulating the shipping process and counting the number of days needed.

Which companies ask LeetCode 1011. Capacity To Ship Packages Within D Days in interviews?

This problem is frequently asked at Agoda, DP world, DoorDash, Expedia, Flipkart and 3 more companies.

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Capacity To Ship Packages Within D Days - LeetCode 1011 Solution

Capacity To Ship Packages Within D Days - Complete Solution Guide

Capacity To Ship Packages Within D Days is LeetCode problem 1011, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A conveyor belt has packages that must be shipped from one port to another within days days. The i th package on the conveyor belt has a weight of weights[i] . Each day, we load the ship with packages on the conveyor belt (in the order given by weights ). We may not load more weight than the maximum weight capacity of the ship. Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within days days. Example 1: Input: weights = [1,2,3,

Detailed Explanation

The problem asks us to find the minimum weight capacity of a ship needed to transport all packages within a given number of days. The packages are represented by an array of weights, and the order in which they are loaded onto the ship matters. We need to find the smallest capacity such that we can ship all packages within the specified number of days. The packages must be loaded in the given order.

Solution Approach

The solution uses binary search to find the minimum capacity. First, the lower and upper bounds for the capacity are determined. The lower bound is the maximum weight in the `weights` array, and the upper bound is the sum of all weights. Then, binary search is performed within this range. For each mid-point (potential capacity), the `can_ship` function is called to check if it's possible to ship all packages within the given number of days with that capacity. If it is, we try a lower capacity (move `high` to `mid`); otherwise, we need a higher capacity (move `low` to `mid + 1`). The binary search continues until `low` and `high` converge, at which point `low` (or `high`) represents the minimum capacity.

Step-by-Step Algorithm

Step 1: Initialize `low` to the maximum weight in the `weights` array and `high` to the sum of all weights.
Step 2: Perform binary search while `low < high`.
Step 3: Calculate the mid-point `mid = low + (high - low) // 2` to avoid overflow.
Step 4: Call the `can_ship` function with `mid` as the capacity to check if it's possible to ship all packages within `days`.
Step 5: If `can_ship(mid)` returns `true`, it means the current capacity is sufficient, so try a smaller capacity by setting `high = mid`.
Step 6: If `can_ship(mid)` returns `false`, it means the current capacity is insufficient, so increase the capacity by setting `low = mid + 1`.
Step 7: After the loop finishes (when `low == high`), `low` will contain the minimum capacity required to ship all packages within `days`.

Key Insights

Insight 1: The problem can be solved using binary search because the possible capacities form a sorted search space. We can define a range of possible capacities (minimum and maximum possible capacity) and iteratively narrow down the search space.
Insight 2: The 'can_ship' function is crucial. It checks whether it's possible to ship all packages within 'days' given a specific capacity. This is done by simulating the shipping process and counting the number of days needed.
Insight 3: The lower bound for the binary search is the maximum weight of a single package, since the ship must be able to carry the heaviest package. The upper bound is the sum of all weights, representing the case where all packages are shipped in a single day.

Complexity Analysis

Time Complexity: O(n*log(sum(weights)))

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search.

Companies

Asked at: Agoda, DP world, DoorDash, Expedia, Flipkart, Mindtickle, Salesforce, Zeta.

Capacity To Ship Packages Within D Days | LeetCode 1011

Medium

Array Binary Search

Time: O(n*log(sum(weights)))Space: O(1)

Solution Code

Available in:

class Solution {
    public int shipWithinDays(int[] weights, int days) {
        int low = 0;
        int high = 0;
        for (int weight : weights) {
            low = Math.max(low, weight);
            high += weight;
        }

        while (low < high) {
            int mid = low + (high - low) / 2;
            if (canShip(weights, days, mid)) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }

        return low;
    }

    private boolean canShip(int[] weights, int days, int capacity) {
        int daysNeeded = 1;
        int currentLoad = 0;
        for (int weight : weights) {
            if (currentLoad + weight > capacity) {
                daysNeeded++;
                currentLoad = weight;
            } else {
                currentLoad += weight;
            }
        }
        return daysNeeded <= days;
    }
}

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Problem Info

Description

Problem 1011: Capacity To Ship Packages Within D Days - Detailed explanation and solution approaches.

Topics

Array Binary Search

Companies

AgodaDP worldDoorDashExpediaFlipkartMindtickleSalesforceZeta

Stats

Difficulty:Medium

Languages:4

Companies:8

Problem #:1011

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

7 steps

1
Step 1: Initialize `low` to the maximum weight in the `weights` array and `high` to the sum of all weights.
2
Step 2: Perform binary search while `low < high`.
3
Step 3: Calculate the mid-point `mid = low + (high - low) // 2` to avoid overflow.
4
Step 4: Call the `can_ship` function with `mid` as the capacity to check if it's possible to ship all packages within `days`.
5
Step 5: If `can_ship(mid)` returns `true`, it means the current capacity is sufficient, so try a smaller capacity by setting `high = mid`.
6
Step 6: If `can_ship(mid)` returns `false`, it means the current capacity is insufficient, so increase the capacity by setting `low = mid + 1`.
7
Step 7: After the loop finishes (when `low == high`), `low` will contain the minimum capacity required to ship all packages within `days`.

💡

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(sum(weights)))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The binary search loop iterates logarithmically with respect to the sum of the weights. Inside each iteration, the can_ship function iterates through the entire weights array linearly. Therefore, the overall time complexity is O(n * log(sum(weights))). The space complexity is constant because no additional data structures dependent on the input size are used.

📖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

Brute Force

Iterate through all possible capacities from max(weights) to sum(weights) and check if each capacity allows shipping within 'days'.

Time: O(n * sum(weights)), where n is the number of weights.Space: O(1)

Trade-offs: This approach is less efficient than binary search because it iterates through all possible capacities linearly. Use binary search as it is significantly faster.

✨

Key Insights

1Insight 1: The problem can be solved using binary search because the possible capacities form a sorted search space. We can define a range of possible capacities (minimum and maximum possible capacity) and iteratively narrow down the search space.
2Insight 2: The 'can_ship' function is crucial. It checks whether it's possible to ship all packages within 'days' given a specific capacity. This is done by simulating the shipping process and counting the number of days needed.
3Insight 3: The lower bound for the binary search is the maximum weight of a single package, since the ship must be able to carry the heaviest package. The upper bound is the sum of all weights, representing the case where all packages are shipped in a single day.

📚

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: `days` is equal to 1. The solution correctly handles this case because the upper bound of the binary search is the sum of all weights, so if only one day is available, the capacity must be the total weight.
Edge case 2: `days` is equal to the number of weights. In this case, each package is shipped on a separate day. The solution correctly handles this, as the lower bound of the binary search is the maximum weight, ensuring that the ship can at least carry the heaviest package.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the `mid` value in the binary search (`mid = (low + high) / 2` can cause overflow for large `low` and `high` values). Use `mid = low + (high - low) // 2` instead.

✗Common mistake 2: Failing to initialize `low` to `max(weights)`. If `low` is initialized to 0, the `can_ship` function will always return `true`, resulting in an infinite loop or incorrect result.

✅

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: Resource allocation - Determining the minimum capacity of trucks to transport goods between warehouses within a specific timeframe.

🏦Real-world scenario 2: Task scheduling - Finding the minimum processing power needed to complete a set of tasks within a deadline, where each task has a certain processing requirement.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: September 20, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

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

@InterviewProJul 6

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

@CodeNinjaJun 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.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Capacity To Ship Packages Within D Days"?

For this medium-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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