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# 1822. Sign of the Product of an Array

🟩 Easy

Implement a function `signFunc(x)` that returns:

* `1` if `x` is positive.
* `-1` if `x` is negative.
* `0` if `x` is equal to `0`. You are given an integer array `nums`. Let `product` be the product of all values in the array `nums`.

Return `signFunc(product)`.

## Example 1

> **Input**: nums = \[-1,-2,-3,-4,3,2,1]\
> **Output**: 2\
> **Explanation**: The product of all values in the array is 144, and signFunc(144) = 1

## Example 2

> **Input**: nums = \[1,5,0,2,-3]\
> **Output**: 0\
> **Explanation**: The product of all values in the array is 0, and signFunc(0) = 0

## Example 3

> **Input**: nums = \[-1,1,-1,1,-1]\
> **Output**: -1\
> **Explanation**: The product of all values in the array is -1, and signFunc(-1) = -1

## Constraints

* `1 <= nums.length <= 1000`
* `-100 <= nums[i] <= 100`

## Hint-1

> If there is a 0 in the array the answer is 0

## Hint-2

> To avoid overflow make all the negative numbers -1 and all positive numbers 1 and calculate the prod

## Solution

My Solution

```go
func arraySign(nums []int) int {
    var sign int8 = 1

    for _, v := range nums {
        if v == 0 {
            return 0
        }

        if v < 0 {
            sign *= -1
        } 
    }
    return int(sign)
}
```

Optimal solution

```go
func arraySign(nums []int) int {
    negativeCount := 0

    // Loop through the array
    for _, num := range nums {
        if num == 0 {
            return 0 // If any number is 0, product will be 0
        }
        if num < 0 {
            negativeCount++ // Count negative numbers
        }
    }

    // If there are an odd number of negative numbers, the product is negative.
    if negativeCount % 2 == 0 {
        return 1 // Positive product
    }
    return -1 // Negative product
}
```

![result](/files/1InQANWU3Nlj1VBqbZ18)

Leetcode: [link](https://leetcode.com/problems/sign-of-the-product-of-an-array/description)


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