LeetCode #3457 — MEDIUM

Eat Pizzas!

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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 that n is a multiple of 4, and each pizza can be eaten only once.

Example 1:

Input: pizzas = [1,2,3,4,5,6,7,8]

Output: 14

Explanation:

On day 1, you eat pizzas at indices [1, 2, 4, 7] = [2, 3, 5, 8]. You gain a weight of 8.
On day 2, you eat pizzas at indices [0, 3, 5, 6] = [1, 4, 6, 7]. You gain a weight of 6.

The total weight gained after eating all the pizzas is 8 + 6 = 14.

Example 2:

Input: pizzas = [2,1,1,1,1,1,1,1]

Output: 3

Explanation:

On day 1, you eat pizzas at indices [4, 5, 6, 0] = [1, 1, 1, 2]. You gain a weight of 2.
On day 2, you eat pizzas at indices [1, 2, 3, 7] = [1, 1, 1, 1]. You gain a weight of 1.

The total weight gained after eating all the pizzas is 2 + 1 = 3.

Constraints:

4 <= n == pizzas.length <= 2 * 10⁵
1 <= pizzas[i] <= 10⁵
n is a multiple of 4.

Step 01

Brute Force Baseline

Problem summary: You are given an integer array pizzas of size n, where pizzas[i] represents the weight of the ith 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 that n is a multiple of 4, and each pizza can be eaten only once.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[1,2,3,4,5,6,7,8]

Example 2

[2,1,1,1,1,1,1,1]

Step 02

Core Insight

What unlocks the optimal approach

On odd-numbered days, it is optimal to pair the smallest three and the largest one.
On even-numbered days, it is optimal to pair the smallest two and the largest two.
There will be ceil((n / 4) / 2) odd-numbered days. Select pizzas for all odd-numbered days first.
Select the remaining pizzas for the even-numbered days.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3457: Eat Pizzas!
class Solution {
    public long maxWeight(int[] pizzas) {
        int n = pizzas.length;
        int days = n / 4;
        Arrays.sort(pizzas);
        int odd = (days + 1) / 2;
        int even = days / 2;
        long ans = 0;
        for (int i = n - odd; i < n; ++i) {
            ans += pizzas[i];
        }
        for (int i = n - odd - 2; even > 0; --even) {
            ans += pizzas[i];
            i -= 2;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3457: Eat Pizzas!
func maxWeight(pizzas []int) (ans int64) {
	n := len(pizzas)
	days := n / 4
	sort.Ints(pizzas)
	odd := (days + 1) / 2
	even := days - odd
	for i := n - odd; i < n; i++ {
		ans += int64(pizzas[i])
	}
	for i := n - odd - 2; even > 0; even-- {
		ans += int64(pizzas[i])
		i -= 2
	}
	return
}

# Accepted solution for LeetCode #3457: Eat Pizzas!
class Solution:
    def maxWeight(self, pizzas: List[int]) -> int:
        days = len(pizzas) // 4
        pizzas.sort()
        odd = (days + 1) // 2
        even = days - odd
        ans = sum(pizzas[-odd:])
        i = len(pizzas) - odd - 2
        for _ in range(even):
            ans += pizzas[i]
            i -= 2
        return ans

// Accepted solution for LeetCode #3457: Eat Pizzas!
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3457: Eat Pizzas!
// class Solution {
//     public long maxWeight(int[] pizzas) {
//         int n = pizzas.length;
//         int days = n / 4;
//         Arrays.sort(pizzas);
//         int odd = (days + 1) / 2;
//         int even = days / 2;
//         long ans = 0;
//         for (int i = n - odd; i < n; ++i) {
//             ans += pizzas[i];
//         }
//         for (int i = n - odd - 2; even > 0; --even) {
//             ans += pizzas[i];
//             i -= 2;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3457: Eat Pizzas!
function maxWeight(pizzas: number[]): number {
    const n = pizzas.length;
    const days = n >> 2;
    pizzas.sort((a, b) => a - b);
    const odd = (days + 1) >> 1;
    let even = days - odd;
    let ans = 0;
    for (let i = n - odd; i < n; ++i) {
        ans += pizzas[i];
    }
    for (let i = n - odd - 2; even; --even) {
        ans += pizzas[i];
        i -= 2;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.