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Formatted question description: https://leetcode.ca/all/2279.html

# 2279. Maximum Bags With Full Capacity of Rocks

• Difficulty: Medium.
• Related Topics: Array, Greedy, Sorting.
• Similar Questions: Capacity To Ship Packages Within D Days, Maximum Units on a Truck.

## Problem

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.**

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

## Solution (Java, C++, Python)

• class Solution {
public int maximumBags(int[] capacity, int[] rocks, int additionalRocks) {
int len = capacity.length;
for (int i = 0; i < len; i++) {
capacity[i] -= rocks[i];
}
Arrays.sort(capacity);
int total = 0;
for (int i = 0; i < len && additionalRocks > 0; i++) {
if (capacity[i] <= additionalRocks) {
total++;
}
}
}
}

############

class Solution {
public int maximumBags(int[] capacity, int[] rocks, int additionalRocks) {
int n = capacity.length;
int[] d = new int[n];
for (int i = 0; i < n; ++i) {
d[i] = capacity[i] - rocks[i];
}
Arrays.sort(d);
int ans = 0;
for (int v : d) {
if (v <= additionalRocks) {
++ans;
} else {
break;
}
}
return ans;
}
}

• class Solution:
def maximumBags(
self, capacity: List[int], rocks: List[int], additionalRocks: int
) -> int:
d = [a - b for a, b in zip(capacity, rocks)]
d.sort()
ans = 0
for v in d:
if v <= additionalRocks:
ans += 1
return ans

############

# 2279. Maximum Bags With Full Capacity of Rocks
# https://leetcode.com/problems/maximum-bags-with-full-capacity-of-rocks/

class Solution:
def maximumBags(self, capacity: List[int], rocks: List[int], extra: int) -> int:
res = 0
A = []

for cap, rock in zip(capacity, rocks):
if cap == rock:
res += 1
continue

A.append(cap - rock)
A.sort()

for x in A:
if extra >= x:
res += 1
extra -= x
else:
break

return res


• class Solution {
public:
int maximumBags(vector<int>& capacity, vector<int>& rocks, int additionalRocks) {
int n = capacity.size();
vector<int> d(n);
for (int i = 0; i < n; ++i) d[i] = capacity[i] - rocks[i];
sort(d.begin(), d.end());
int ans = 0;
for (int& v : d) {
if (v > additionalRocks) break;
++ans;
}
return ans;
}
};

• func maximumBags(capacity []int, rocks []int, additionalRocks int) int {
n := len(capacity)
d := make([]int, n)
for i, v := range capacity {
d[i] = v - rocks[i]
}
sort.Ints(d)
ans := 0
for _, v := range d {
if v > additionalRocks {
break
}
ans++
}
return ans
}

• function maximumBags(
capacity: number[],
rocks: number[],
): number {
const n = capacity.length;
const diffs = capacity.map((c, i) => c - rocks[i]);
diffs.sort((a, b) => a - b);
let ans = 0;
for (
let i = 0;
i < n && (diffs[i] === 0 || diffs[i] <= additionalRocks);
i++
) {
ans++;
}
return ans;
}


• impl Solution {
pub fn maximum_bags(capacity: Vec<i32>, rocks: Vec<i32>, mut additional_rocks: i32) -> i32 {
let n = capacity.len();
let mut diffs = vec![0; n];
for i in 0..n {
diffs[i] = capacity[i] - rocks[i];
}
diffs.sort();
for i in 0..n {
if diffs[i] > additional_rocks {
return i as i32;
}
}
n as i32
}
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).