100% Faster🔥100% Lesser🔥4 Methods🔥Brute Force🔥Using Counter🔥Math Library🔥Bit Manipulation
Here are 4 ways to solve this question with detailed explanation of each approach:\n\n# 1. Brute Force:\n\n- Iterate through each number in the array.\n- If the
# Here are 4 ways to solve this question with detailed explanation of each approach:\n\n# 1. Brute Force:\n\n- Iterate through each number in the array.\n- If the number is 0, return 0 as the product will be 0.\n- If the number is negative, toggle a flag variable indicating the presence of negative numbers.\n- After iterating through all the numbers, if the flag is set, return -1; otherwise, return 1.\n# Complexity\n- Time complexity: O(n), where n is the length of the array.\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# Code\n```Python []\nclass Solution:\n    def arraySign(self, nums: List[int]) -> int:\n        product = 1\n        for num in nums:\n            if num == 0:\n                return 0\n            product *= num\n        if product > 0:\n            return 1\n        elif product < 0:\n            return -1\n        else:\n            return 0\n\n```\n```Java []\nclass Solution {\n    public int arraySign(int[] nums) {\n        int product = 1;\n        for (int num : nums) {\n            if (num == 0) {\n                return 0;\n            }\n            product *= num;\n        }\n        if (product > 0) {\n            return 1;\n        } else if (product < 0) {\n            return -1;\n        } else {\n            return 0;\n        }\n    }\n}\n\n```\n```C++ []\nclass Solution {\npublic:\n    int arraySign(vector<int>& nums) {\n        int product = 1;\n        for (int num : nums) {\n            if (num == 0) {\n                return 0;\n            }\n            product *= num;\n        }\n        if (product > 0) {\n            return 1;\n        } else if (product < 0) {\n            return -1;\n        } else {\n            return 0;\n        }\n    }\n};\n\n```\n```C []\nint arraySign(int* nums, int numsSize) {\n    int product = 1;\n    for (int i = 0; i < numsSize; i++) {\n        if (nums[i] == 0) {\n            return 0;\n        }\n        product *= nums[i];\n    }\n    if (product > 0) {\n        return 1;\n    } else if (product < 0) {\n        return -1;\n    } else {\n        return 0;\n    }\n}\n\n```\n# 2. Using a Counter:\n   - Initialize a counter variable to 0.\n   - Iterate through each number in the array.\n   - If the number is 0, return 0 as the product will be 0.\n   - If the number is negative, increment the counter.\n   - After iterating through all the numbers, check the value of the counter.\n   - If it is an even number, return 1; otherwise, return -1.\n# Complexity\n- Time complexity: O(n), where n is the length of the array.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- Space complexity: O(1).\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n# Code\n```Python []\nclass Solution:\n    def arraySign(self, nums: List[int]) -> int:\n        counter = 0\n        for num in nums:\n            if num == 0:\n                return 0\n            elif num < 0:\n                counter += 1\n        return 1 if counter % 2 == 0 else -1\n```\n```Java []\nclass Solution {\n    public int arraySign(int[] nums) {\n        int counter = 0;\n        for (int num : nums) {\n            if (num == 0) {\n                return 0;\n            } else if (num < 0) {\n                counter++;\n            }\n        }\n        return (counter % 2 == 0) ? 1 : -1;\n    }\n}\n```\n```C++ []\nclass Solution {\npublic:\n    int arraySign(std::vector<int>& nums) {\n        int counter = 0;\n        for (int num : nums) {\n            if (num == 0) {\n                return 0;\n            } else if (num < 0) {\n                counter++;\n            }\n        }\n        return (counter % 2 == 0) ? 1 : -1;\n    }\n};\n```\n```C []\nint arraySign(int* nums, int numsSize) {\n    int counter = 0;\n    for (int i = 0; i < numsSize; i++) {\n        if (nums[i] == 0) {\n            return 0;\n        } else if (nums[i] < 0) {\n            counter++;\n        }\n    }\n    return (counter % 2 == 0) ? 1 : -1;\n}\n```\n# 3. Math Library:\n   - Import the math module.\n   - Initialize a product variable to 1.\n   - Iterate through each number in the array.\n   - Multiply the product by the number.\n   - After iterating through all the numbers, use the math library\'s sign function (math.prod) to determine the sign of the product.\n   - Return the sign.\n# Complexity\n- Time complexity: O(n), where n is the length of the array.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- Space complexity: O(1).\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n# Code\n```Python []\nclass Solution:\n    def arraySign(self, nums: List[int]) -> int:\n        product = math.prod(nums)\n        return 1 if product > 0 else -1 if product < 0 else 0\n```\n```Java []\nclass Solution {\n    public int arraySign(int[] nums) {\n        int product = Arrays.stream(nums).reduce(1, (a, b) -> a * b);\n        return Integer.compare(product, 0);\n    }\n}\n```\n```C++ []\nclass Solution {\npublic:\n    int arraySign(vector<int>& nums) {\n        int product = accumulate(nums.begin(), nums.end(), 1, multiplies<int>());\n        return (product > 0) ? 1 : (product < 0) ? -1 : 0;\n    }\n};\n```\n```C []\nint arraySign(int* nums, int numsSize) {\n    int product = 1;\n    for (int i = 0; i < numsSize; i++) {\n        product *= nums[i];\n    }\n    return (product > 0) ? 1 : (product < 0) ? -1 : 0;\n}\n\n```\n# 4. Bit Manipulation:\n   - Initialize a variable called sign with a value of 1.\n   - Initialize a variable called zero_count with a value of 0\n   - Iterate through each number in the array.\n   - If the number is 0,increment zero_count.\n   - If the number is negative, toggle the sign variable.\n   - After iterating through all the numbers, check the value of zero_count.\n   - If zero_count is greater than 0, return 0 as the product will be 0.\n   - Otherwise, return the value of sign.\n   \n# Complexity\n- Time complexity: O(n), where n is the length of the array.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- Space complexity: O(1).\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n# Code\n```Python []\nclass Solution:\n    def arraySign(self, nums: List[int]) -> int:\n        sign = 1\n        zero_count = 0\n        \n        for num in nums:\n            if num == 0:\n                return 0\n            elif num < 0:\n                sign ^= 1\n\n        return sign\n```\n```Java []\nclass Solution {\n    public int arraySign(int[] nums) {\n        int sign = 1;\n        int zeroCount = 0;\n        \n        for (int num : nums) {\n            if (num == 0) {\n                return 0;\n            } else if (num < 0) {\n                sign ^= 1;\n            }\n        }\n        \n        return sign;\n    }\n}\n\n```\n```C++ []\nclass Solution {\npublic:\n    int arraySign(std::vector<int>& nums) {\n        int sign = 1;\n        int zeroCount = 0;\n        \n        for (int num : nums) {\n            if (num == 0) {\n                return 0;\n            } else if (num < 0) {\n                sign ^= 1;\n            }\n        }\n        \n        return sign;\n    }\n};\n```\n```C []\nint arraySign(int* nums, int numsSize) {\n    int sign = 1;\n    int zeroCount = 0;\n    \n    for (int i = 0; i < numsSize; i++) {\n        if (nums[i] == 0) {\n            return 0;\n        } else if (nums[i] < 0) {\n            sign ^= 1;\n        }\n    }\n    \n    return sign;\n}\n```\n# Note: \nThe most efficient way to solve the "Sign of the Product of an Array" problem is to use a single pass approach without actually calculating the product of the array.\n\nThe time and space complexities mentioned above are based on the algorithmic approach used in each solution. The actual time and space requirements may vary depending on the language and specific implementation.
['Math', 'C', 'C++', 'Java', 'Python3']