LeetCode #2279 — MEDIUM

Maximum Bags With Full Capacity of Rocks

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

The Problem

Problem Statement

You have n bags numbered from 0 to n - 1. You are given two 0-indexed integer arrays capacity and rocks. The i^th bag can hold a maximum of capacity[i] rocks and currently contains rocks[i] rocks. You are also given an integer additionalRocks, the number of additional rocks you can place in any of the bags.

Return the maximum number of bags that could have full capacity after placing the additional rocks in some bags.

Example 1:

Input: capacity = [2,3,4,5], rocks = [1,2,4,4], additionalRocks = 2
Output: 3
Explanation:
Place 1 rock in bag 0 and 1 rock in bag 1.
The number of rocks in each bag are now [2,3,4,4].
Bags 0, 1, and 2 have full capacity.
There are 3 bags at full capacity, so we return 3.
It can be shown that it is not possible to have more than 3 bags at full capacity.
Note that there may be other ways of placing the rocks that result in an answer of 3.

Example 2:

Input: capacity = [10,2,2], rocks = [2,2,0], additionalRocks = 100
Output: 3
Explanation:
Place 8 rocks in bag 0 and 2 rocks in bag 2.
The number of rocks in each bag are now [10,2,2].
Bags 0, 1, and 2 have full capacity.
There are 3 bags at full capacity, so we return 3.
It can be shown that it is not possible to have more than 3 bags at full capacity.
Note that we did not use all of the additional rocks.

Constraints:

n == capacity.length == rocks.length
1 <= n <= 5 * 10⁴
1 <= capacity[i] <= 10⁹
0 <= rocks[i] <= capacity[i]
1 <= additionalRocks <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You have n bags numbered from 0 to n - 1. You are given two 0-indexed integer arrays capacity and rocks. The ith bag can hold a maximum of capacity[i] rocks and currently contains rocks[i] rocks. You are also given an integer additionalRocks, the number of additional rocks you can place in any of the bags. Return the maximum number of bags that could have full capacity after placing the additional rocks in some bags.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

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

Example 2

[10,2,2]
[2,2,0]
100

Core Insight

What unlocks the optimal approach

Which bag should you fill completely first?
Can you think of a greedy solution?

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 #2279: Maximum Bags With Full Capacity of Rocks
class Solution {
    public int maximumBags(int[] capacity, int[] rocks, int additionalRocks) {
        int n = rocks.length;
        for (int i = 0; i < n; ++i) {
            capacity[i] -= rocks[i];
        }
        Arrays.sort(capacity);
        for (int i = 0; i < n; ++i) {
            additionalRocks -= capacity[i];
            if (additionalRocks < 0) {
                return i;
            }
        }
        return n;
    }
}

// Accepted solution for LeetCode #2279: Maximum Bags With Full Capacity of Rocks
func maximumBags(capacity []int, rocks []int, additionalRocks int) int {
	for i, x := range rocks {
		capacity[i] -= x
	}
	sort.Ints(capacity)
	for i, x := range capacity {
		additionalRocks -= x
		if additionalRocks < 0 {
			return i
		}
	}
	return len(capacity)
}

# Accepted solution for LeetCode #2279: Maximum Bags With Full Capacity of Rocks
class Solution:
    def maximumBags(
        self, capacity: List[int], rocks: List[int], additionalRocks: int
    ) -> int:
        for i, x in enumerate(rocks):
            capacity[i] -= x
        capacity.sort()
        for i, x in enumerate(capacity):
            additionalRocks -= x
            if additionalRocks < 0:
                return i
        return len(capacity)

// Accepted solution for LeetCode #2279: Maximum Bags With Full Capacity of Rocks
impl Solution {
    pub fn maximum_bags(mut capacity: Vec<i32>, rocks: Vec<i32>, mut additional_rocks: i32) -> i32 {
        for i in 0..rocks.len() {
            capacity[i] -= rocks[i];
        }
        capacity.sort();
        for i in 0..capacity.len() {
            additional_rocks -= capacity[i];
            if additional_rocks < 0 {
                return i as i32;
            }
        }
        capacity.len() as i32
    }
}

// Accepted solution for LeetCode #2279: Maximum Bags With Full Capacity of Rocks
function maximumBags(capacity: number[], rocks: number[], additionalRocks: number): number {
    const n = rocks.length;
    for (let i = 0; i < n; ++i) {
        capacity[i] -= rocks[i];
    }
    capacity.sort((a, b) => a - b);
    for (let i = 0; i < n; ++i) {
        additionalRocks -= capacity[i];
        if (additionalRocks < 0) {
            return i;
        }
    }
    return n;
}

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.