### Problem

We need to find a maximum product of 2 numbers in a provided vector of numbers `vec`

. In math, *a product of 2 numbers* is a result of multiplication of these 2 numbers.

#### Example 1:

*Input:* `vec = [-10, -3, 5, 7, -2]`

*Output:* `35`

*Explanation:* A product of `(5, 7)`

is the max product.

#### Example 2:

*Input:* `vec = [-10, -3, 5, 4, -2]`

*Output:* `30`

*Explanation:* A product of `(-10, -3)`

is the max product.

#### Example 3:

*Input:* `vec = [-5, 0, 10]`

*Output:* `0`

*Explanation:* A product of `(-5, 0)`

or `(0, 10)`

is the max product.

### Solution

```
fn find_max_product_of_2_numbers(vec: &Vec<i32>) -> i32 {
let mut max_pos_1 = 0;
let mut max_pos_2 = 0;
let mut min_neg_1 = 0;
let mut min_neg_2 = 0;
for e in vec {
if *e > 0 && *e > max_pos_1 {
max_pos_2 = max_pos_1;
max_pos_1 = *e;
} else if *e > 0 && *e > max_pos_2 {
max_pos_2 = *e;
}
if *e < 0 && *e < min_neg_1 {
min_neg_2 = min_neg_1;
min_neg_1 = *e;
} else if *e < 0 && *e < min_neg_2 {
min_neg_2 = *e;
}
}
let product_of_max_positives = max_pos_1 * max_pos_2;
let product_of_min_negatives = min_neg_1 * min_neg_2;
match (product_of_max_positives, product_of_min_negatives) {
(pos, neg) if pos != 0 && neg != 0 && pos >= neg => pos, //product of positives is larger
(pos, neg) if pos != 0 && neg != 0 && pos <= neg => neg, //product of negatives is larger
(pos, 0) if pos != 0 => pos, //has only a product of positives
(0, neg) if neg != 0 => neg, //has only a product of negatives
_ if max_pos_1 != 0 && min_neg_1 != 0 && vec.len() == 2 => max_pos_1 * min_neg_1, //has 1 pos & 1 neg number
_ => 0, //otherwise returns 0
}
}
```

### Explanation

In general, a maximum product number would come from either multiplying 2 maximum positive numbers OR 2 minimum negative numbers. It means that we need to iterate once over the vector `vec`

to find these 4 numbers (`max_pos_1`

, `max_pos_2`

, `min_neg_1`

, `min_neg_2`

). Then we could find a product of 2 maximum positive numbers AND a product of 2 minimum negative numbers. A maximum value from these products would give us a maximum product overall.

Also we need to handle special cases when there are no positive (or negative) pairs in the vector `vec`

, or when the vector contains zeros.

### Complexity

The time complexity of this solution is `O(n)`

(`n`

is a size of the input vector).