What is the time complexity of Put Marbles in Bags?

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

What data structures are used to solve Put Marbles in Bags?

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

What are the key insights for solving Put Marbles in Bags?

Insight 1: The sum of the first and last elements of the original `weights` array is always included in the total score, regardless of how the marbles are divided into bags. Therefore, optimizing the score involves maximizing or minimizing the sums of weights at the 'cut points' where bags are split. Insight 2: The problem can be reframed to finding the `k-1` largest and `k-1` smallest sums of adjacent weights, and calculating the difference of their sums, since each cut contributes the sum of the weights immediately before and after that cut.

Which companies ask LeetCode 2551. Put Marbles in Bags in interviews?

This problem is frequently asked at DE Shaw, Flipkart.

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Put Marbles in Bags - LeetCode 2551 Solution

Put Marbles in Bags - Complete Solution Guide

Put Marbles in Bags is LeetCode problem 2551, 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 k bags. You are given a 0-indexed integer array weights where weights[i] is the weight of the i th marble. You are also given the integer k. Divide the marbles into the k bags according to the following rules: No bag is empty. If the i th marble and j th marble are in a bag, then all marbles with an index between the i th and j th indices should also be in that same bag. If a bag consists of all the marbles with an index from i to j inclusively, then the cost of the bag is weights[i] +

Detailed Explanation

The problem asks us to divide an array of marble weights into `k` bags. The cost of each bag is the sum of the weights of the first and last marble in the bag. The goal is to find the difference between the maximum and minimum possible total cost across all valid divisions. A valid division must have no empty bags, and if marbles `i` and `j` are in the same bag, all marbles between them must also be in the same bag. The core challenge is to efficiently find the maximum and minimum scores achievable.

Solution Approach

The solution computes the sums of adjacent weights in the `weights` array and stores these in a new array called `pair_sums`. This `pair_sums` array is then sorted. To maximize the total score, we select the `k-1` largest elements from the sorted `pair_sums` and sum them. To minimize the total score, we select the `k-1` smallest elements from the sorted `pair_sums` and sum them. Finally, we return the difference between these two sums.

Step-by-Step Algorithm

Step 1: Handle the base cases where `k` is 1 or `k` is equal to the length of `weights`. In these cases, return 0.
Step 2: Create a `pair_sums` array to store the sums of adjacent elements in the `weights` array. For each `i` from 0 to `n-2`, calculate `weights[i] + weights[i+1]` and store it in `pair_sums[i]`.
Step 3: Sort the `pair_sums` array in ascending order.
Step 4: Calculate the minimum score sum by summing the first `k-1` elements of the sorted `pair_sums` array.
Step 5: Calculate the maximum score sum by summing the last `k-1` elements of the sorted `pair_sums` array.
Step 6: Return the difference between the maximum score sum and the minimum score sum.

Key Insights

Insight 1: The sum of the first and last elements of the original `weights` array is always included in the total score, regardless of how the marbles are divided into bags. Therefore, optimizing the score involves maximizing or minimizing the sums of weights at the 'cut points' where bags are split.
Insight 2: The problem can be reframed to finding the `k-1` largest and `k-1` smallest sums of adjacent weights, and calculating the difference of their sums, since each cut contributes the sum of the weights immediately before and after that cut.
Insight 3: The trivial cases of `k = 1` or `k = n` result in a difference of 0, because there's only one possible distribution.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: DE Shaw, Flipkart.

Put Marbles in Bags | LeetCode 2551

Hard

Array Greedy Sorting Heap (Priority Queue)

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public long putMarbles(int[] weights, int k) {
        int n = weights.length;
        if (k == 1 || k == n) {
            return 0;
        }

        // The total score can be expressed as:
        // score = weights[0] + weights[n-1] + sum(weights[i] + weights[i+1])
        // for each of the k-1 chosen cut points i.
        //
        // To maximize the score, we choose the k-1 largest `weights[i] + weights[i+1]`.
        // To minimize the score, we choose the k-1 smallest `weights[i] + weights[i+1]`.
        //
        // The difference between max and min score is therefore:
        // (sum of k-1 largest pair sums) - (sum of k-1 smallest pair sums).

        int[] pair_sums = new int[n - 1];
        for (int i = 0; i < n - 1; i++) {
            pair_sums[i] = weights[i] + weights[i+1];
        }
        
        Arrays.sort(pair_sums);
        
        long min_score_sum = 0;
        long max_score_sum = 0;
        
        for (int i = 0; i < k - 1; i++) {
            min_score_sum += pair_sums[i];
            max_score_sum += pair_sums[n - 2 - i];
        }
        
        return max_score_sum - min_score_sum;
    }
}

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Problem Info

Description

Problem 2551: Put Marbles in Bags - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting Heap (Priority Queue)

Companies

DE ShawFlipkart

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2551

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

6 steps

1
Step 1: Handle the base cases where `k` is 1 or `k` is equal to the length of `weights`. In these cases, return 0.
2
Step 2: Create a `pair_sums` array to store the sums of adjacent elements in the `weights` array. For each `i` from 0 to `n-2`, calculate `weights[i] + weights[i+1]` and store it in `pair_sums[i]`.
3
Step 3: Sort the `pair_sums` array in ascending order.
4
Step 4: Calculate the minimum score sum by summing the first `k-1` elements of the sorted `pair_sums` array.
5
Step 5: Calculate the maximum score sum by summing the last `k-1` elements of the sorted `pair_sums` array.
6
Step 6: Return the difference between the maximum score sum and the minimum score sum.

💡

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 `pair_sums` array requires O(n) space. The sorting algorithm used by `Arrays.sort()` in Java, `std::sort()` in C++, and `list.sort()` in Python usually has a space complexity of O(log n) or O(n) depending on the implementation, but it is dominated by the space needed for `pair_sums` itself.

📖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

Heap (Priority Queue)

Instead of sorting the `pair_sums` array, we could use a min-heap (priority queue) of size `k-1` to find the `k-1` smallest sums, and a max-heap of size `k-1` to find the `k-1` largest sums. This could potentially be more efficient if `k` is much smaller than `n`.

Time: O(n log k), where n is the length of the `weights` array and k is the number of bags.Space: O(k), since we store up to `k-1` elements in each heap.

Trade-offs: This approach is better when `k` is much smaller than `n`. If `k` is close to `n`, then sorting might be faster. The constant factors involved in heap operations might also affect performance.

✨

Key Insights

1Insight 1: The sum of the first and last elements of the original `weights` array is always included in the total score, regardless of how the marbles are divided into bags. Therefore, optimizing the score involves maximizing or minimizing the sums of weights at the 'cut points' where bags are split.
2Insight 2: The problem can be reframed to finding the `k-1` largest and `k-1` smallest sums of adjacent weights, and calculating the difference of their sums, since each cut contributes the sum of the weights immediately before and after that cut.
3Insight 3: The trivial cases of `k = 1` or `k = n` result in a difference of 0, because there's only one possible distribution.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

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

Edge Cases

Edge case 1: `k = 1` or `k = n`. In these cases, there's only one way to distribute the marbles, so the minimum and maximum scores are the same, and the difference is 0. The code handles this at the beginning by returning 0.
Edge case 2: `n = 1`. Although not explicitly covered in the problem description, an array with only one weight should not result in a `pair_sums` calculation. The given constraints say 1 <= k <= weights.length, so n=1 implies k=1 which is handled by the base case return 0
Edge case 3: When `k` is close to `n`. The code correctly handles the case where we need to sum almost all of the pair sums. For example, `k=n-1`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the edge cases where `k = 1` or `k = n`, which result in a 0 difference.

✗Common mistake 2: Forgetting that the first and last elements of `weights` are always part of the total score, and focusing solely on the adjacent pair sums. Though this doesn't affect the final difference calculation, understanding this helps clarify the problem.

✗Common mistake 3: Incorrectly calculating the number of elements to sum from the sorted `pair_sums` array (using `k` instead of `k-1`).

✗Common mistake 4: Integer overflow issues with large input weights; using `long` or `long long` where appropriate.

✅

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 the placement of advertisements on a webpage to maximize click-through rate. Similar to cutting the `weights` array, the ad placement affects the viewer's perception. By choosing the right places to make 'cuts', we can maximize clicks based on the individual ad performances at each place.

🏦Real-world scenario 2: Resource allocation. Imagine dividing a series of tasks among `k` workers. Each task has a 'weight' representing its complexity or time required. The cost of assigning a range of tasks to a worker could be the sum of the first and last task's weights. The goal is to minimize the total cost of assigning tasks to workers, similar to minimizing the total score in this problem.

💡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: July 2, 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.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

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 "Put Marbles in Bags"?

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

🤖 Need help with this problem?