Eat Pizzas!

Eat Pizzas! - Problem

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 (the heaviest pizza).

On even-numbered days, you gain a weight of Y (the second heaviest pizza).

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.

Input & Output

Example 1 — Single Day

$ Input: pizzas = [3, 1, 4, 2]

› Output: 4

💡 Note: We have 4 pizzas, so exactly 1 day of eating. Day 1 is odd, so we gain the weight of the heaviest pizza. Sorting [3,1,4,2] gives [1,2,3,4], so we gain weight 4.

Example 2 — Multiple Days

$ Input: pizzas = [10, 8, 7, 6, 5, 4, 3, 2]

› Output: 25

💡 Note: We have 8 pizzas = 2 days. Day 1 (odd): gain heaviest from group = 10. Day 2 (even): gain second heaviest from group = 8. Optimal grouping gives us 10 + 8 + 7 = 25.

Example 3 — Equal Weights

$ Input: pizzas = [5, 5, 5, 5]

› Output: 5

💡 Note: All pizzas have equal weight. Day 1 (odd): any grouping will have Z = 5, so we gain 5. Total weight gained is 5.

Constraints

4 ≤ pizzas.length ≤ 1000
pizzas.length is a multiple of 4
1 ≤ pizzas[i] ≤ 1000

Visualization

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Asked in

G Google 42 f Meta 38 a Amazon 35 M Microsoft 28

The key insight is that we want to maximize weight gained by strategically assigning our heaviest pizzas to positions that contribute to our total. Since odd days take the heaviest pizza (Z) and even days take second heaviest (Y), we should sort pizzas and assign the top pizzas to these contributing positions. Best approach uses greedy sorting. Time: O(n log n), Space: O(1).

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Brute Force - Try All Combinations

⏱️ Time: O(C(n,4) × 2^(n/4)) Space: O(n/4)

Generate all possible ways to partition pizzas into groups of 4, then try all possible assignments of these groups to odd and even days. For each assignment, calculate the total weight gained.

Greedy Sorting Approach

⏱️ Time: O(n log n) Space: O(1)

Sort all pizzas by weight, then greedily assign the heaviest pizzas to positions where they contribute most (Z position on odd days, Y position on even days), while pairing them with lightest available pizzas.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

⚡ Linearithmic Space

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Smart Actions

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