# 1822. Sign of the Product of an Array

## Description

There is 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: 1
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

## Solutions

Solution 1: Direct Traversal

The problem requires us to return the sign of the product of the array elements, i.e., return $1$ for positive numbers, $-1$ for negative numbers, and $0$ if it equals $0$.

We can define an answer variable ans, initially set to $1$.

Then we traverse each element $v$ in the array. If $v$ is a negative number, we multiply ans by $-1$. If $v$ is $0$, we return $0$ in advance.

After the traversal is over, we return ans.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

• class Solution {
public int arraySign(int[] nums) {
int ans = 1;
for (int v : nums) {
if (v == 0) {
return 0;
}
if (v < 0) {
ans *= -1;
}
}
return ans;
}
}

• class Solution {
public:
int arraySign(vector<int>& nums) {
int ans = 1;
for (int v : nums) {
if (!v) return 0;
if (v < 0) ans *= -1;
}
return ans;
}
};

• class Solution:
def arraySign(self, nums: List[int]) -> int:
ans = 1
for v in nums:
if v == 0:
return 0
if v < 0:
ans *= -1
return ans


• func arraySign(nums []int) int {
ans := 1
for _, v := range nums {
if v == 0 {
return 0
}
if v < 0 {
ans *= -1
}
}
return ans
}

• /**
* @param {number[]} nums
* @return {number}
*/
var arraySign = function (nums) {
let ans = 1;
for (const v of nums) {
if (!v) {
return 0;
}
if (v < 0) {
ans *= -1;
}
}
return ans;
};


• impl Solution {
pub fn array_sign(nums: Vec<i32>) -> i32 {
let mut ans = 1;
for &num in nums.iter() {
if num == 0 {
return 0;
}
if num < 0 {
ans *= -1;
}
}
ans
}
}