Here's an explanation of the code:

class Solution {
public:
    int arraySign(vector<int>& nums) {
        int countNeg = 0;
        for (int num: nums) {
            if (num == 0) return 0;
            if (num < 0) countNeg++;
        }
        return (countNeg % 2 == 0) ? 1 : -1;
    }
};

1. Initialize countNeg to 0, which will store the count of negative numbers.
2. Iterate through each number num in the input array nums.
3. If num is 0, return 0 immediately, since any number multiplied by 0 is 0.
4. If num is negative, increment countNeg.
5. After iterating through all numbers, check if countNeg is even (i.e., countNeg % 2 == 0).
• If even, return 1, since an even number of negative numbers will result in a positive product.
• If odd, return -1, since an odd number of negative numbers will result in a negative product.

Example 1:

Example 2:

Example 3:

Example 4:

class Solution {
public:

    int signFunc(int x) {
    return (x > 0) ? 1 : (x < 0) ? -1 : 0;
    }
  
    int arraySign(vector<int>& nums) {
        int result = 1 ; 
        for(auto n: nums){
            result *= signFunc(n) ; 
        }
        return result ; 
    }
};

if (x > 0) {
    return 1;
} else if (x < 0) {
    return -1;
} else {
    return 0;
}