**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

You have `n`

bags numbered from `0`

to `n - 1`

. You are given two **0-indexed** integer arrays `capacity`

and `rocks`

. The `i`

bag can hold a maximum of ^{th}`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 * 10`

^{4}`1 <= capacity[i] <= 10`

^{9}`0 <= rocks[i] <= capacity[i]`

`1 <= additionalRocks <= 10`

^{9}

```
public int maximumBags(int[] capacity, int[] rocks, int additionalRocks) {
List<Integer> list = new ArrayList<>();
for(int i=0; i<rocks.length; i++){
list.add(capacity[i]-rocks[i]);
}
Collections.sort(list);
int ans=0,count=0;
for(int a:list){
count += a;
if(count > additionalRocks) break;
ans++;
}
return ans;
}
```

```
int maximumBags(vector<int>& c, vector<int>& r, int additionalRocks) {
transform(begin(c), end(c), begin(r), begin(c), [](int c, int r){ return c - r; });
sort(begin(c), end(c));
int i = 0;
for (; i < c.size() && c[i] <= additionalRocks; ++i)
additionalRocks -= c[i];
return i;
}
```

```
def maximumBags(self, capacity, rocks, x):
count = sorted(c - r for c,r in zip(capacity, rocks))[::-1]
while count and x and count[-1] <= x:
x -= count.pop()
return len(rocks) - len(count)
```

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