## Problem – Make Array Zero by Subtracting Equal Amounts LeetCode Solution

You are given a non-negative integer array `nums`

. In one operation, you must:

- Choose a positive integer
`x`

such that `x`

is less than or equal to the **smallest non-zero** element in `nums`

. - Subtract
`x`

from every **positive** element in `nums`

.

Return *the ***minimum** number of operations to make every element in `nums`

* equal to *`0`

.

**Example 1:**

```
Input: nums = [1,5,0,3,5]
Output: 3
Explanation:
In the first operation, choose x = 1. Now, nums = [0,4,0,2,4].
In the second operation, choose x = 2. Now, nums = [0,2,0,0,2].
In the third operation, choose x = 2. Now, nums = [0,0,0,0,0].
```

**Example 2:**

```
Input: nums = [0]
Output: 0
Explanation: Each element in nums is already 0 so no operations are needed.
```

**Constraints:**

`1 <= nums.length <= 100`

`0 <= nums[i] <= 100`

### Make Array Zero by Subtracting Equal Amounts LeetCode Solution in Java

```
public int minimumOperations(int[] nums) {
Set<Integer> set = new HashSet<>();
for (int a: nums)
if (a > 0)
set.add(a);
return set.size();
}
```

### Make Array Zero by Subtracting Equal Amounts LeetCode Solution in C++

```
int minimumOperations(vector<int>& A) {
unordered_set<int> s = unordered_set(A.begin(), A.end());
return s.size() - s.count(0);
}
```

### Make Array Zero by Subtracting Equal Amounts LeetCode Solution in Python

```
def minimumOperations(self, nums):
return len(set(x for x in nums if x > 0))
```

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