LeetCode #3074 — EASY

Apple Redistribution into Boxes

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given an array apple of size n and an array capacity of size m.

There are n packs where the i^th pack contains apple[i] apples. There are m boxes as well, and the i^th box has a capacity of capacity[i] apples.

Return the minimum number of boxes you need to select to redistribute these n packs of apples into boxes.

Note that, apples from the same pack can be distributed into different boxes.

Example 1:

Input: apple = [1,3,2], capacity = [4,3,1,5,2]
Output: 2
Explanation: We will use boxes with capacities 4 and 5.
It is possible to distribute the apples as the total capacity is greater than or equal to the total number of apples.

Example 2:

Input: apple = [5,5,5], capacity = [2,4,2,7]
Output: 4
Explanation: We will need to use all the boxes.

Constraints:

1 <= n == apple.length <= 50
1 <= m == capacity.length <= 50
1 <= apple[i], capacity[i] <= 50
The input is generated such that it's possible to redistribute packs of apples into boxes.

Step 01

Brute Force Baseline

Problem summary: You are given an array apple of size n and an array capacity of size m. There are n packs where the ith pack contains apple[i] apples. There are m boxes as well, and the ith box has a capacity of capacity[i] apples. Return the minimum number of boxes you need to select to redistribute these n packs of apples into boxes. Note that, apples from the same pack can be distributed into different boxes.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[1,3,2]
[4,3,1,5,2]

Example 2

[5,5,5]
[2,4,2,7]

Step 02

Core Insight

What unlocks the optimal approach

Sort array <code>capacity</code> in non-decreasing order.
Greedily select boxes with the largest capacities to redistribute apples optimally.

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 #3074: Apple Redistribution into Boxes
class Solution {
    public int minimumBoxes(int[] apple, int[] capacity) {
        Arrays.sort(capacity);
        int s = 0;
        for (int x : apple) {
            s += x;
        }
        for (int i = 1, n = capacity.length;; ++i) {
            s -= capacity[n - i];
            if (s <= 0) {
                return i;
            }
        }
    }
}

// Accepted solution for LeetCode #3074: Apple Redistribution into Boxes
func minimumBoxes(apple []int, capacity []int) int {
	sort.Ints(capacity)
	s := 0
	for _, x := range apple {
		s += x
	}
	for i := 1; ; i++ {
		s -= capacity[len(capacity)-i]
		if s <= 0 {
			return i
		}
	}
}

# Accepted solution for LeetCode #3074: Apple Redistribution into Boxes
class Solution:
    def minimumBoxes(self, apple: List[int], capacity: List[int]) -> int:
        capacity.sort(reverse=True)
        s = sum(apple)
        for i, c in enumerate(capacity, 1):
            s -= c
            if s <= 0:
                return i

// Accepted solution for LeetCode #3074: Apple Redistribution into Boxes
impl Solution {
    pub fn minimum_boxes(apple: Vec<i32>, mut capacity: Vec<i32>) -> i32 {
        capacity.sort();

        let mut s: i32 = apple.iter().sum();

        let n = capacity.len();
        let mut i = 1;
        loop {
            s -= capacity[n - i];
            if s <= 0 {
                return i as i32;
            }
            i += 1;
        }
    }
}

// Accepted solution for LeetCode #3074: Apple Redistribution into Boxes
function minimumBoxes(apple: number[], capacity: number[]): number {
    capacity.sort((a, b) => b - a);
    let s = apple.reduce((acc, cur) => acc + cur, 0);
    for (let i = 1; ; ++i) {
        s -= capacity[i - 1];
        if (s <= 0) {
            return i;
        }
    }
}

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(m × log m + n)

Space

O(log m)

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.