What is the time complexity of Maximum Spending After Buying Items?

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

What data structures are used to solve Maximum Spending After Buying Items?

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

What are the key insights for solving Maximum Spending After Buying Items?

Insight 1: Greedy approach is suitable because buying the smallest available item each day from all shops ensures that the larger valued items are reserved for later days, maximizing their contribution to the total spending due to the increasing day multiplier. Insight 2: A min-heap (priority queue) is ideal for efficiently tracking the smallest available item across all shops and updating the choices as items are bought.

Which companies ask LeetCode 2931. Maximum Spending After Buying Items in interviews?

This problem is frequently asked at Zomato.

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Maximum Spending After Buying Items - LeetCode 2931 Solution

Q: How to solve Maximum Spending After Buying Items?

The solution uses a greedy approach combined with a min-heap. We initialize the min-heap with the last (smallest) item from each shop. On each day, we extract the smallest item from the heap, add its cost (value * day) to the total spending, and then, if there are more items in the shop from which the item was extracted, add the next smallest item from that shop to the heap.

Maximum Spending After Buying Items - Complete Solution Guide

Maximum Spending After Buying Items is LeetCode problem 2931, 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 are given a 0-indexed m * n integer matrix values , representing the values of m * n different items in m different shops. Each shop has n items where the j th item in the i th shop has a value of values[i][j] . Additionally, the items in the i th shop are sorted in non-increasing order of value. That is, values[i][j] >= values[i][j + 1] for all 0 <= j < n - 1 . On each day, you would like to buy a single item from one of the shops. Specifically, On the d th day you can: Pick any shop i . Bu

Detailed Explanation

The problem asks us to maximize the total spending by buying items from 'm' shops, each containing 'n' items sorted in non-increasing order of value. We can buy only one item each day. The cost of an item on day 'd' is its value multiplied by 'd'. The goal is to determine the optimal order of item purchases to maximize the total spending across all m*n days.

Solution Approach

The solution uses a greedy approach combined with a min-heap. We initialize the min-heap with the last (smallest) item from each shop. On each day, we extract the smallest item from the heap, add its cost (value * day) to the total spending, and then, if there are more items in the shop from which the item was extracted, add the next smallest item from that shop to the heap.

Step-by-Step Algorithm

Step 1: Initialize a min-heap (priority queue) to store tuples of (item_value, shop_index, item_index) for the last item from each shop. Since the items in each shop are sorted in non-increasing order, the last item has the smallest value.
Step 2: Initialize 'total_spending' to 0 and 'day' to 1.
Step 3: While the min-heap is not empty:
Step 4: Extract the minimum element (val, shop_idx, item_idx) from the min-heap.
Step 5: Update total_spending: total_spending += val * day.
Step 6: Increment day: day += 1.
Step 7: If the shop at shop_idx has more items (item_idx > 0), get the value of the previous item (values[shop_idx][item_idx - 1]) and insert (values[shop_idx][item_idx - 1], shop_idx, item_idx - 1) into the min-heap.
Step 8: Return total_spending.

Key Insights

Insight 1: Greedy approach is suitable because buying the smallest available item each day from all shops ensures that the larger valued items are reserved for later days, maximizing their contribution to the total spending due to the increasing day multiplier.
Insight 2: A min-heap (priority queue) is ideal for efficiently tracking the smallest available item across all shops and updating the choices as items are bought.
Insight 3: The fact that each shop's items are sorted in non-increasing order simplifies the process of finding the 'rightmost available' item in each shop, as we can maintain a pointer or index for each shop that tracks the last unbought item.

Complexity Analysis

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

Space Complexity: O(m)

Topics

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

Companies

Asked at: Zomato.

Maximum Spending After Buying Items | LeetCode 2931

Hard

Array Greedy Sorting Heap (Priority Queue)Matrix

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

Solution Code

Available in:

import java.util.PriorityQueue;

class Solution {
    public long maxSpending(int[][] values) {
        int m = values.length;
        int n = values[0].length;
        
        PriorityQueue<int[]> min_heap = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        for (int i = 0; i < m; i++) {
            min_heap.offer(new int[]{values[i][n - 1], i, n - 1});
        }
        
        long total_spending = 0;
        long day = 1;
        
        while (!min_heap.isEmpty()) {
            int[] curr = min_heap.poll();
            int val = curr[0];
            int shop_idx = curr[1];
            int item_idx = curr[2];
            
            total_spending += (long)val * day;
            day++;
            
            if (item_idx > 0) {
                int next_item_idx = item_idx - 1;
                int next_val = values[shop_idx][next_item_idx];
                min_heap.offer(new int[]{next_val, shop_idx, next_item_idx});
            }
        }
        
        return total_spending;
    }
}

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Problem Info

Description

Problem 2931: Maximum Spending After Buying Items - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting Heap (Priority Queue)Matrix

Companies

Zomato

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2931

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

8 steps

1
Step 1: Initialize a min-heap (priority queue) to store tuples of (item_value, shop_index, item_index) for the last item from each shop. Since the items in each shop are sorted in non-increasing order, the last item has the smallest value.
2
Step 2: Initialize 'total_spending' to 0 and 'day' to 1.
3
Step 3: While the min-heap is not empty:
4
Step 4: Extract the minimum element (val, shop_idx, item_idx) from the min-heap.
5
Step 5: Update total_spending: total_spending += val * day.
6
Step 6: Increment day: day += 1.
7
Step 7: If the shop at shop_idx has more items (item_idx > 0), get the value of the previous item (values[shop_idx][item_idx - 1]) and insert (values[shop_idx][item_idx - 1], shop_idx, item_idx - 1) into the min-heap.
8
Step 8: Return total_spending.

💡

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(m*n*log(m))

💾

Space Complexity

O(m)

📝

Why?

The time complexity is determined by the (m*n) iterations, each having O(log(m)) heap operations. The heap's size depends on the number of shops, hence the space complexity of O(m).

📖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

Sorting Approach

Extract all values into a single array, multiply each value by an index, sort it and the sum all the values.

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

Trade-offs: This appproach has higher time and space complexity because it needs to store all values, whereas the minHeap approach keeps track of the minimum and only holds at most 'm' elements.

✨

Key Insights

1Insight 1: Greedy approach is suitable because buying the smallest available item each day from all shops ensures that the larger valued items are reserved for later days, maximizing their contribution to the total spending due to the increasing day multiplier.
2Insight 2: A min-heap (priority queue) is ideal for efficiently tracking the smallest available item across all shops and updating the choices as items are bought.
3Insight 3: The fact that each shop's items are sorted in non-increasing order simplifies the process of finding the 'rightmost available' item in each shop, as we can maintain a pointer or index for each shop that tracks the last unbought item.

📚

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: Empty values array. While the constraints state m >= 1 and n >= 1, a check might be implemented for robustness. If the input array is empty, the code returns 0.
Edge case 2: A shop with only one item. The code correctly handles this case because after the item is extracted from the heap, the 'if item_idx > 0' condition will be false, so no more items from that shop will be added to the heap.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a min-heap. Using a simple array and searching for the minimum each time would lead to O(m) time complexity for each extraction, making the overall solution O(m^2*n), which exceeds the time limit.

✗Common mistake 2: Incorrectly implementing the heap. A faulty heap implementation could lead to incorrect ordering and, therefore, a suboptimal solution.

✅

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: Task scheduling where tasks have different priorities and deadlines. We want to maximize the value of completing tasks by scheduling high-priority tasks closer to their deadlines.

🏦Real-world scenario 2: Resource allocation in cloud computing where we need to allocate resources to different users with varying demands to maximize overall system utilization and 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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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 22, 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 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

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 "Maximum Spending After Buying Items"?

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.

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