What is the time complexity of Eat Pizzas!?

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

What data structures are used to solve Eat Pizzas!?

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

What are the key insights for solving Eat Pizzas!?

Insight 1: Sorting the pizzas allows us to easily identify the heaviest pizzas. Insight 2: Eating the heaviest pizzas on odd days and the next heaviest on even days maximizes the total weight gained because heaviest pizza weight is considered on odd days.

Which companies ask LeetCode 3457. Eat Pizzas! in interviews?

This problem is frequently asked at Infosys.

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Eat Pizzas! - LeetCode 3457 Solution

Q: How to solve Eat Pizzas!?

The solution involves sorting the pizza weights in ascending order. Then, we calculate the number of odd and even days. For odd days, we eat the largest pizzas. For even days, we select the largest available pizzas, avoiding overlapping with already eaten largest pizzas for odd days. The total weight gained is the sum of weights from odd and even days.

Eat Pizzas! - Complete Solution Guide

Eat Pizzas! is LeetCode problem 3457, 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 are given an integer array pizzas of size n , where pizzas[i] represents the weight of the i th pizza. Every day, you eat exactly 4 pizzas. Due to your incredible metabolism, when you eat pizzas of weights W , X , Y , and Z , where W <= X <= Y <= Z , you gain the weight of only 1 pizza! On odd-numbered days (1-indexed) , you gain a weight of Z . On even-numbered days, you gain a weight of Y . Find the maximum total weight you can gain by eating all pizzas optimally. Note : It is guaranteed t

Detailed Explanation

The problem asks us to maximize the total weight gained from eating pizzas. We are given an array of pizza weights and must eat exactly 4 pizzas each day. The weight gained each day is equal to the second largest weight (on even days) or the largest weight (on odd days) among the four pizzas eaten that day. The goal is to find the optimal way to eat all pizzas to maximize the total weight gained, given that the number of pizzas is a multiple of 4.

Solution Approach

The solution involves sorting the pizza weights in ascending order. Then, we calculate the number of odd and even days. For odd days, we eat the largest pizzas. For even days, we select the largest available pizzas, avoiding overlapping with already eaten largest pizzas for odd days. The total weight gained is the sum of weights from odd and even days.

Step-by-Step Algorithm

Step 1: Sort the `pizzas` array in ascending order.
Step 2: Calculate the number of days `k` required to eat all pizzas: `k = n / 4` where `n` is the number of pizzas.
Step 3: Calculate the number of odd days: `k_odd = (k + 1) // 2`.
Step 4: Calculate the number of even days: `k_even = k // 2`.
Step 5: Calculate the weight gained on odd days by summing the `k_odd` largest pizza weights: These are the pizzas at the end of the sorted array (from `n - k_odd` to `n`).
Step 6: Calculate the weight gained on even days by summing the `k_even` largest pizza weights *after* removing the pizzas eaten on odd days: This consists of the pizzas from index `n - k - k_even` to `n - k`.
Step 7: Return the sum of the weight gained on odd days and the weight gained on even days.

Key Insights

Insight 1: Sorting the pizzas allows us to easily identify the heaviest pizzas.
Insight 2: Eating the heaviest pizzas on odd days and the next heaviest on even days maximizes the total weight gained because heaviest pizza weight is considered on odd days.
Insight 3: Determining the optimal number of odd and even days to plan the pizza consumption.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy, Sorting.

Companies

Asked at: Infosys.

Eat Pizzas! | LeetCode 3457

Medium

Array Greedy Sorting

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public long maxWeight(int[] pizzas) {
        Arrays.sort(pizzas);
        int n = pizzas.length;
        int k = n / 4;

        int k_odd = (k + 1) / 2;
        int k_even = k / 2;

        long total_weight = 0;

        // Gain from odd days.
        // This corresponds to the sum of the k_odd largest pizzas.
        if (k_odd > 0) {
            for (int i = n - k_odd; i < n; i++) {
                total_weight += pizzas[i];
            }
        }

        // Gain from even days.
        // This corresponds to the sum of k_even pizzas from a specific range
        // of the remaining pizzas. The indices in the sorted array are
        // from n - k - k_even to n - k - 1.
        if (k_even > 0) {
            int start_index = n - k - k_even;
            int end_index = n - k;
            for (int i = start_index; i < end_index; i++) {
                total_weight += pizzas[i];
            }
        }

        return total_weight;
    }
}

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Problem Info

Description

Problem 3457: Eat Pizzas! - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting

Companies

Infosys

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3457

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: Sort the `pizzas` array in ascending order.
2
Step 2: Calculate the number of days `k` required to eat all pizzas: `k = n / 4` where `n` is the number of pizzas.
3
Step 3: Calculate the number of odd days: `k_odd = (k + 1) // 2`.
4
Step 4: Calculate the number of even days: `k_even = k // 2`.
5
Step 5: Calculate the weight gained on odd days by summing the `k_odd` largest pizza weights: These are the pizzas at the end of the sorted array (from `n - k_odd` to `n`).
6
Step 6: Calculate the weight gained on even days by summing the `k_even` largest pizza weights *after* removing the pizzas eaten on odd days: This consists of the pizzas from index `n - k - k_even` to `n - k`.
7
Step 7: Return the sum of the weight gained on odd days and the weight gained on even 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 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 the array of pizzas which takes O(n log n) time. The rest of the operations, such as calculating the number of odd/even days and summing the weights, take O(n) time in the worst case. Since O(n log n) dominates O(n), the overall time complexity is O(n log n). The space complexity is O(1) because we are sorting in-place and using a fixed number of variables.

📖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

Using a Max Heap

Instead of sorting, maintain a max-heap of size k. Repeatedly pop 4 elements and determine the largest or second largest based on whether it's an odd or even day.

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

Trade-offs: This could be more efficient if k is significantly smaller than n, but requires additional memory for the heap.

✨

Key Insights

1Insight 1: Sorting the pizzas allows us to easily identify the heaviest pizzas.
2Insight 2: Eating the heaviest pizzas on odd days and the next heaviest on even days maximizes the total weight gained because heaviest pizza weight is considered on odd days.
3Insight 3: Determining the optimal number of odd and even days to plan the pizza consumption.

📚

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: `n = 4`. In this case, `k = 1`, `k_odd = 1`, and `k_even = 0`. We take the largest pizza only on the first (odd) day and the correct result is calculated.
Edge case 2: `n = 8`. In this case, `k = 2`, `k_odd = 1`, and `k_even = 1`. We take the largest pizza on the first (odd) day and the next largest pizza on the second (even) day. Again, this is handled correctly by the algorithm.
Edge case 3: All pizzas have the same weight. The algorithm will still function correctly as sorting will not affect the order and we will always take the correct number of pizzas.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the pizzas first. This leads to incorrect selection of the heaviest pizzas for each day.

✗Common mistake 2: Incorrectly calculating the indices for the 'even day' pizzas. It's crucial to understand how the 'even day' pizzas are chosen from the remaining pizzas after accounting for the 'odd day' pizzas.

✗Common mistake 3: Using integer division when calculating the sums. While the problem states integers are used for pizza weight and size, summing many pizza weights can lead to overflow. The languages where appropriate type promotion should be implemented (using long type for sum).

✅

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. Optimizing the allocation of resources with varying values, where a limited number of resources can be used each time.

🏦Real-world scenario 2: Task scheduling. Scheduling tasks to maximize output based on certain tasks providing larger rewards on specific days.

💡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: July 8, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

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

@TechLeadJun 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.

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 "Eat Pizzas!"?

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

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