What is the time complexity of Minimum Space Wasted From Packaging?

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

What data structures are used to solve Minimum Space Wasted From Packaging?

This problem uses Array, Binary Search, Sorting, Prefix Sum. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Space Wasted From Packaging?

Insight 1: Sorting both packages and boxes allows for efficient searching (binary search) to find the smallest box that can fit a particular package. Insight 2: Iterating through each supplier's boxes and calculating the wasted space independently allows us to compare the results and find the minimum wasted space.

Which companies ask LeetCode 1889. Minimum Space Wasted From Packaging in interviews?

This problem is frequently asked at Two Sigma.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Minimum Space Wasted From Packaging - LeetCode 1889 Solution

Minimum Space Wasted From Packaging - Complete Solution Guide

Minimum Space Wasted From Packaging is LeetCode problem 1889, 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 n packages that you are trying to place in boxes, one package in each box . There are m suppliers that each produce boxes of different sizes (with infinite supply). A package can be placed in a box if the size of the package is less than or equal to the size of the box. The package sizes are given as an integer array packages , where packages[i] is the size of the i th package. The suppliers are given as a 2D integer array boxes , where boxes[j] is an array of box sizes that the j th su

Detailed Explanation

The problem requires finding the minimum wasted space when fitting packages into boxes provided by different suppliers. Each package must go into exactly one box. Wasted space is defined as the difference between the box size and the package size. We need to choose one supplier such that the total wasted space across all packages is minimized. If no supplier can accommodate all packages, we return -1. The final result should be returned modulo 10^9 + 7.

Solution Approach

The solution iterates through each supplier, sorts their boxes, and then attempts to fit the packages into the boxes. For each supplier, the packages are iterated through (implicitly via binary search). The binary search identifies the optimal box size for fitting the packages while calculating the total wasted space. The minimum wasted space across all suppliers is tracked and updated, returning -1 if no supplier can fit all packages. The final result is modulo 10^9 + 7.

Step-by-Step Algorithm

Step 1: Sort the `packages` array to enable efficient searching.
Step 2: Iterate through each `supplier_boxes` array in the `boxes` array.
Step 3: Sort the `supplier_boxes` array.
Step 4: Check if the largest box size in `supplier_boxes` is smaller than the largest package size in `packages`. If so, skip to the next supplier (optimization).
Step 5: Initialize `current_total_box_space` and `last_package_idx` to 0.
Step 6: Iterate through each `box_size` in `supplier_boxes`.
Step 7: Use binary search (bisect_right in Python, binarySearchRight in Java, upper_bound in C++) to find the index of the first package that cannot fit into `box_size` starting from `last_package_idx`.
Step 8: Calculate the number of packages that can fit in the current `box_size` as `idx - last_package_idx`. Update `current_total_box_space` by adding `(num_packages_for_this_box * box_size)`.
Step 9: Update `last_package_idx` to `idx`.
Step 10: If all packages have been assigned a box (`last_package_idx == num_packages`), break out of the inner loop.
Step 11: Calculate `current_waste` as `current_total_box_space - total_package_size`.
Step 12: Update `min_total_waste` with the minimum between the current `min_total_waste` and `current_waste`.
Step 13: After iterating through all suppliers, if `min_total_waste` remains at its initial value (infinity), return -1. Otherwise, return `min_total_waste % MOD`.

Key Insights

Insight 1: Sorting both packages and boxes allows for efficient searching (binary search) to find the smallest box that can fit a particular package.
Insight 2: Iterating through each supplier's boxes and calculating the wasted space independently allows us to compare the results and find the minimum wasted space.
Insight 3: Checking if the largest box of a supplier is smaller than the largest package is crucial for early exit, as it indicates that supplier cannot accommodate all packages.

Complexity Analysis

Time Complexity: O(n log n + m * (k log k + n log n))

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Sorting, Prefix Sum.

Companies

Asked at: Two Sigma.

Minimum Space Wasted From Packaging | LeetCode 1889

Hard

Array Binary Search Sorting Prefix Sum

Time: O(n log n + m * (k log k + n log n))Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int minWastedSpace(int[] packages, int[][] boxes) {
        int MOD = 1000000007;

        Arrays.sort(packages);
        int numPackages = packages.length;
        long totalPackageSize = 0;
        for (int packageSize : packages) {
            totalPackageSize += packageSize;
        }

        long minTotalWaste = Long.MAX_VALUE;

        for (int[] supplierBoxes : boxes) {
            Arrays.sort(supplierBoxes);

            if (supplierBoxes[supplierBoxes.length - 1] < packages[packages.length - 1]) {
                continue;
            }

            long currentTotalBoxSpace = 0;
            int lastPackageIdx = 0;

            for (int boxSize : supplierBoxes) {
                int idx = binarySearchRight(packages, boxSize, lastPackageIdx);

                int numPackagesForThisBox = idx - lastPackageIdx;
                currentTotalBoxSpace += (long)numPackagesForThisBox * boxSize;

                lastPackageIdx = idx;

                if (lastPackageIdx == numPackages) {
                    break;
                }
            }

            long currentWaste = currentTotalBoxSpace - totalPackageSize;
            minTotalWaste = Math.min(minTotalWaste, currentWaste);
        }

        if (minTotalWaste == Long.MAX_VALUE) {
            return -1;
        } else {
            return (int)(minTotalWaste % MOD);
        }
    }

    private int binarySearchRight(int[] arr, int target, int lo) {
        int hi = arr.length;
        while (lo < hi) {
            int mid = lo + (hi - lo) / 2;
            if (arr[mid] <= target) {
                lo = mid + 1;
            } else {
                hi = mid;
            }
        }
        return lo;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1889: Minimum Space Wasted From Packaging - Detailed explanation and solution approaches.

Topics

Array Binary Search Sorting Prefix Sum

Companies

Two Sigma

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1889

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

13 steps

1
Step 1: Sort the `packages` array to enable efficient searching.
2
Step 2: Iterate through each `supplier_boxes` array in the `boxes` array.
3
Step 3: Sort the `supplier_boxes` array.
4
Step 4: Check if the largest box size in `supplier_boxes` is smaller than the largest package size in `packages`. If so, skip to the next supplier (optimization).
5
Step 5: Initialize `current_total_box_space` and `last_package_idx` to 0.
6
Step 6: Iterate through each `box_size` in `supplier_boxes`.
7
Step 7: Use binary search (bisect_right in Python, binarySearchRight in Java, upper_bound in C++) to find the index of the first package that cannot fit into `box_size` starting from `last_package_idx`.
8
Step 8: Calculate the number of packages that can fit in the current `box_size` as `idx - last_package_idx`. Update `current_total_box_space` by adding `(num_packages_for_this_box * box_size)`.
9
Step 9: Update `last_package_idx` to `idx`.
10
Step 10: If all packages have been assigned a box (`last_package_idx == num_packages`), break out of the inner loop.
11
Step 11: Calculate `current_waste` as `current_total_box_space - total_package_size`.
12
Step 12: Update `min_total_waste` with the minimum between the current `min_total_waste` and `current_waste`.
13
Step 13: After iterating through all suppliers, if `min_total_waste` remains at its initial value (infinity), return -1. Otherwise, return `min_total_waste % MOD`.

💡

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 + m * (k log k + n log n))

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

Sorting packages takes O(n log n) time. The outer loop iterates through each supplier. For each supplier, sorting their boxes takes O(k log k) time, where k is the number of boxes the supplier provides. The binary search inside the inner loop contributes O(n log n) in the worst case. The space complexity is O(1) because we are not using any additional data structures that scale with the input size, aside from a few variables to keep track of indices and sums.

📖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

While not efficient, one could explore a DP approach where dp[i] represents the minimum wasted space using boxes up to index i. This would require considering all possible subsets of boxes, making it less efficient than the current solution.

Time: Exponential in the worst caseSpace: O(n)

Trade-offs: DP is generally not suitable for this problem due to its high time complexity. It might be useful in variations with stricter constraints.

✨

Key Insights

1Insight 1: Sorting both packages and boxes allows for efficient searching (binary search) to find the smallest box that can fit a particular package.
2Insight 2: Iterating through each supplier's boxes and calculating the wasted space independently allows us to compare the results and find the minimum wasted space.
3Insight 3: Checking if the largest box of a supplier is smaller than the largest package is crucial for early exit, as it indicates that supplier cannot accommodate all packages.

📚

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!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: When a supplier cannot fit all packages because its largest box is smaller than the largest package. The solution handles this by skipping the supplier using the condition `if supplier_boxes[-1] < packages[-1]: continue`.
Edge case 2: When no supplier can fit all packages. The solution handles this by checking if `min_total_waste` remains at its initial maximum value (infinity). If so, it returns -1.
Edge case 3: Empty input arrays. The code should work fine with empty boxes arrays as the loop will simply not execute. If the packages array is empty, then the total_package_size will be 0 and the min_total_waste will also be 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the packages and boxes, which would make the binary search inefficient or impossible.

✗Common mistake 2: Integer overflow when calculating the total wasted space. Using long long (or long in Java) is important to prevent overflow, especially with large input sizes. Remember to cast multiplication of `int` to `long` before assigning to a `long` variable.

✗Common mistake 3: Failing to handle the case where no supplier can fit all the packages, which should result in returning -1.

✗Common mistake 4: Forgetting to take the modulo of the result (10^9 + 7), as required by the problem statement.

✅

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 resource allocation in cloud computing, where VMs (boxes) of different sizes are used to run applications (packages). Minimizing wasted space can lead to cost savings.

🏦Real-world scenario 2: Inventory management, where different sized items (packages) need to be stored in containers (boxes). Choosing the right container size to minimize wasted space reduces storage costs.

💡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: August 29, 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

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

@AlgoMasterJun 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.

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 "Minimum Space Wasted From Packaging"?

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.

🤖 Need help with this problem?