What is the time complexity of Sell Diminishing-Valued Colored Balls?

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

What data structures are used to solve Sell Diminishing-Valued Colored Balls?

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

What are the key insights for solving Sell Diminishing-Valued Colored Balls?

Insight 1: Sorting the inventory in descending order allows us to prioritize selling colors with the highest initial value. Insight 2: The problem can be approached greedily by selling balls from colors with higher counts first. Calculate how many full blocks of sales can occur before moving on.

Which companies ask LeetCode 1648. Sell Diminishing-Valued Colored Balls in interviews?

This problem is frequently asked at Groupon, MathWorks.

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Sell Diminishing-Valued Colored Balls - LeetCode 1648 Solution

Sell Diminishing-Valued Colored Balls - Complete Solution Guide

Sell Diminishing-Valued Colored Balls is LeetCode problem 1648, 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

You have an inventory of different colored balls, and there is a customer that wants orders balls of any color. The customer weirdly values the colored balls. Each colored ball's value is the number of balls of that color you currently have in your inventory . For example, if you own 6 yellow balls, the customer would pay 6 for the first yellow ball. After the transaction, there are only 5 yellow balls left, so the next yellow ball is then valued at 5 (i.e., the value of the balls decreases as y

Detailed Explanation

The problem requires us to maximize the total value obtained by selling a certain number of colored balls (orders) from an inventory. Each color has a specific number of balls initially. The value of a ball of a particular color is equal to the number of balls of that color currently available in the inventory. The value decreases as we sell more balls of the same color. The goal is to find the maximum total value achievable after selling the required number of balls, modulo 10^9 + 7.

Solution Approach

The provided code uses a greedy approach combined with the arithmetic series formula to calculate the maximum profit. It sorts the inventory in descending order and then iterates through the inventory. In each iteration, it calculates the difference between the current color's value and the next color's value. It determines whether it can sell all balls within this difference for each color, or if it needs to sell a partial amount from the current level of the inventory. The calculations are performed with modulo to avoid integer overflow.

Step-by-Step Algorithm

Step 1: Sort the inventory array in descending order.
Step 2: Append 0 to the inventory array. This handles the case where the last color needs to be completely sold.
Step 3: Initialize `ans` (total profit) and `width` (number of colors having the same count) to 0.
Step 4: Iterate through the sorted inventory (except the last element):
Step 5: Increment `width` for each consecutive element.
Step 6: Check if the current inventory value is greater than the next inventory value.
Step 7: If the condition in step 6 is true, calculate the difference between the current and next inventory values.
Step 8: Calculate the maximum number of balls that can be sold (`can_sell`) based on `width` and the difference.
Step 9: If `orders` is greater than or equal to `can_sell`, sell all possible balls from the current level, update `ans` using the arithmetic series formula, and reduce `orders`. Use modulo operation after summing.
Step 10: If `orders` is less than `can_sell`, sell a partial amount from the current level, calculate the remaining balls to sell, and update `ans`. The arithmetic series is again used, along with a correction for the remainder. Return `ans` at the end.
Step 11: Modulo operation (%) is used to keep the value of the total profit within the required range.

Key Insights

Insight 1: Sorting the inventory in descending order allows us to prioritize selling colors with the highest initial value.
Insight 2: The problem can be approached greedily by selling balls from colors with higher counts first. Calculate how many full blocks of sales can occur before moving on.
Insight 3: The sum of an arithmetic series formula can efficiently compute the value obtained from selling a range of balls.
Insight 4: The modulo operation is crucial to avoid integer overflow, as the values can get very large.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

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

Companies

Asked at: Groupon, MathWorks.

Sell Diminishing-Valued Colored Balls | LeetCode 1648

Medium

Array Math Binary Search Greedy Sorting Heap (Priority Queue)

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int maxProfit(int[] inventory, int orders) {
        int MOD = 1000000007;

        Arrays.sort(inventory);
        int[] temp = new int[inventory.length + 1];
        for (int i = 0; i < inventory.length; i++) {
            temp[i] = inventory[i];
        }
        temp[inventory.length] = 0;
        inventory = temp;

        // Reverse the array
        for (int i = 0, j = inventory.length - 1; i < j; i++, j--) {
            int tmp = inventory[i];
            inventory[i] = inventory[j];
            inventory[j] = tmp;
        }
        

        int n = inventory.length;

        long ans = 0;
        long width = 0;

        for (int i = 0; i < n - 1; i++) {
            width += 1;
            if (inventory[i] > inventory[i+1]) {
                long current_val = inventory[i];
                long next_val = inventory[i+1];
                long diff = current_val - next_val;

                long can_sell = width * diff;

                if (orders >= can_sell) {
                    // Sell all balls from current_val down to next_val + 1
                    // Sum of arithmetic series: sum(1..b) - sum(1..a-1)
                    long val_sum = (current_val * (current_val + 1) / 2) - (next_val * (next_val + 1) / 2);
                    ans = (ans + width * val_sum) % MOD;
                    orders -= can_sell;
                } else {
                    // Sell a partial amount from the current level
                    long full_rounds = orders / width;
                    long final_val = current_val - full_rounds;

                    // Profit from selling 'full_rounds' from each of the 'width' piles
                    long val_sum = (current_val * (current_val + 1) / 2) - (final_val * (final_val + 1) / 2);
                    ans = (ans + width * val_sum) % MOD;

                    // Profit from the remainder
                    long remainder = orders % width;
                    ans = (ans + remainder * final_val) % MOD;

                    // All orders fulfilled
                    return (int) ans;
                }
            }
        }
        
        return (int) ans;
    }
}

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Problem Info

Description

Problem 1648: Sell Diminishing-Valued Colored Balls - Detailed explanation and solution approaches.

Topics

Array Math Binary Search Greedy Sorting Heap (Priority Queue)

Companies

GrouponMathWorks

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1648

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

11 steps

1
Step 1: Sort the inventory array in descending order.
2
Step 2: Append 0 to the inventory array. This handles the case where the last color needs to be completely sold.
3
Step 3: Initialize `ans` (total profit) and `width` (number of colors having the same count) to 0.
4
Step 4: Iterate through the sorted inventory (except the last element):
5
Step 5: Increment `width` for each consecutive element.
6
Step 6: Check if the current inventory value is greater than the next inventory value.
7
Step 7: If the condition in step 6 is true, calculate the difference between the current and next inventory values.
8
Step 8: Calculate the maximum number of balls that can be sold (`can_sell`) based on `width` and the difference.
9
Step 9: If `orders` is greater than or equal to `can_sell`, sell all possible balls from the current level, update `ans` using the arithmetic series formula, and reduce `orders`. Use modulo operation after summing.
10
Step 10: If `orders` is less than `can_sell`, sell a partial amount from the current level, calculate the remaining balls to sell, and update `ans`. The arithmetic series is again used, along with a correction for the remainder. Return `ans` at the end.
11
Step 11: Modulo operation (%) is used to keep the value of the total profit within the required range.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation is sorting, which has a time complexity of O(n log n). The other operations, such as iterating through the array and performing calculations, take linear time O(n). However, O(n log n) is greater than O(n), so the overall time complexity is O(n log n). The space complexity is O(1) because it only uses a few variables to store intermediate results. The java and C solutions allocate O(N) but this can be avoided.

📖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

Binary Search

Binary search can be used to find the optimal minimum value 'x' to sell each item to. The sum from initial value to x for each item is calculated to determine whether the total orders can be met. This requires more complex logic and can be harder to understand.

Time: O(n log m), where n is the number of items in the inventory and m is the maximum value in the inventory.Space: O(1)

Trade-offs: Binary search may be faster for very large inventories and small order sizes. However, the arithmetic series based approach is easier to implement and understand and often performs competitively.

Priority Queue (Heap)

A priority queue (max heap) can store the inventory values. Repeatedly, the largest value is extracted, added to the total profit, and reduced by one, then re-inserted back into the heap, up to 'orders' times. The heap ensures that you sell the balls with the highest current value first.

Time: O(orders * log n), where n is the number of items in the inventory. 'orders' times we extract the max value (log n) and re-insert. This can be inefficient when orders are large.Space: O(n), for storing the inventory values in the heap.

Trade-offs: This approach is intuitive but less efficient if orders are very large. Slower than arithmetic series and binary search.

✨

Key Insights

1Insight 1: Sorting the inventory in descending order allows us to prioritize selling colors with the highest initial value.
2Insight 2: The problem can be approached greedily by selling balls from colors with higher counts first. Calculate how many full blocks of sales can occur before moving on.
3Insight 3: The sum of an arithmetic series formula can efficiently compute the value obtained from selling a range of balls.
4Insight 4: The modulo operation is crucial to avoid integer overflow, as the values can get very large.

📚

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!

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: inventory = [10], orders = 5. The code correctly calculates the profit by selling 5 balls with values 10, 9, 8, 7, 6.
Edge case 2: orders exceeding the total inventory. If orders is larger than the sum of the initial inventory, the solution will correctly calculate the max possible value. In this case, it will sell all balls.
Edge case 3: Empty inventory. While the constraints state the inventory will never be empty, it is always good practice to check, since the algorithm adds an element, accessing the first element without checking size can cause an error.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. Forgetting to use the modulo operator can cause the values to exceed the maximum integer range and lead to incorrect results.

✗Common mistake 2: Not sorting the inventory. Without sorting, the greedy approach will not work correctly, and the maximum profit won't be achieved.

✗Common mistake 3: Incorrectly calculating the arithmetic series sum. Ensure the sum (n * (n + 1)) / 2 is calculated correctly and use long long to avoid overflow.

✗Common mistake 4: Misunderstanding the modulo operator and applying it incorrectly. Be certain the modulo operator is applied after each summation.

✅

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. Imagine you are managing a cloud computing platform and need to allocate resources (CPU, memory) to different users. Each resource has a diminishing value as more users consume it. This algorithm helps maximize the value you derive from allocating these resources.

🏦Real-world scenario 2: Dynamic pricing. A store wants to sell a limited number of products with decreasing prices depending on the remaining quantity. This solution can help the store to decide the optimal pricing strategy to maximize revenue.

🎮Real-world scenario 3: Ad placement. An advertising platform has a limited number of ad slots available. Each ad slot's value decreases as more ads are placed. The algorithm helps to find the optimal ad placement strategy to maximize revenue.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 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

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

@CodeNinjaJul 13

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

@TechLeadJul 6

❓

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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Sell Diminishing-Valued Colored Balls"?

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

🤖 Need help with this problem?