Simple Python3 Solution, 0(n) time, 0(1) space
\n1. If nums has 0 it means the result of multiplication is guranteed to be 0\n2. Otherwise count the number of negative signs\n3. If the number of negative sig
```\n1. If nums has 0 it means the result of multiplication is guranteed to be 0\n2. Otherwise count the number of negative signs\n3. If the number of negative signs is even it means they are all cancelled during multiplication and we can return 1 else -1\n\n\nclass Solution:\n    def arraySign(self, nums: List[int]) -> int:\n        if 0 in nums:\n            return 0\n        \n        count = 0\n        \n        for n in nums:\n            if n < 0:\n                count += 1\n        \n        return 1 if count % 2 == 0 else -1\n```
['Python', 'Python3']