## Maximum Bags With Full Capacity of Rocks solution leetcode

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:**

## Maximum Bags With Full Capacity of Rocks solution leetcode

Input:capacity = [2,3,4,5], rocks = [1,2,4,4], additionalRocks = 2Output:3Explanation: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:**

## Maximum Bags With Full Capacity of Rocks solution leetcode

Input:capacity = [10,2,2], rocks = [2,2,0], additionalRocks = 100Output:3Explanation: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.

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**Constraints:**

## Maximum Bags With Full Capacity of Rocks solution leetcode

`n == capacity.length == rocks.length`

`1 <= n <= 5 * 10`

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

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

`1 <= additionalRocks <= 10`

^{9}

## SOLUTION

## “Click here“