Python||Single pass approach||100% Faster||Space Complexity: O(1)||Time Complexity: O(n)||Easy
Intuition\n Describe your first thoughts on how to solve this problem. \nCounting the number of negative integers in the array. If the count is odd, the product
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nCounting the number of negative integers in the array. If the count is odd, the product is negative, otherwise, it is positive. This approach has a time complexity of O(n), where n is the length of the array, and it avoids the issue of integer overflow.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe approach to solve the "Sign of the Product of an Array" problem efficiently is to count the number of negative integers in the array. The sign of the product of the array depends on the number of negative integers.\n\nIf the number of negative integers is even, then the product of the array is positive. If the number of negative integers is odd, then the product of the array is negative. If there are no negative integers in the array, then the product of the array is positive.\n\nSo, to solve the problem efficiently, we can traverse the array once and count the number of negative integers. If the count is even, we return 1 (positive sign). If the count is odd, we return -1 (negative sign). If there are no negative integers, we return 0 (positive sign). This approach has a time complexity of O(N), where N is the length of the array.\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def arraySign(self, nums: List[int]) -> int:\n        neg_count = 0\n        for num in nums:\n            if num == 0:\n                return 0\n            elif num < 0:\n                neg_count += 1\n        return -1 if neg_count % 2 == 1 else 1\n```
['Math', 'Counting', 'Python', 'Python3']