JAVA EASIEST SOLUTION || Using ONly IF and ELSE
Approach\nCount the number of negative integer, if number of count is divisible by 2 that means it will be a *positive value so the answer is going to be 1, if 
# Approach\nCount the number of **negative integer**, if number of count is **divisible by 2 **that means it will be a ****positive value**** so the answer is going to be 1, if the number of count is** not divisibleby 2 **that means it will be a **negative number**. and if the array contains any 0 so the product is ultimately going to 0,so the answer is 0. \n\n\n# Code\n```\nclass Solution {\n    public int arraySign(int[] nums) {\n        int count = 0,product = 1,result = 0;\n        for (int num : nums) {\n            if (num < 0) count++;\n            if (num == 0) {\n                product = 0;\n                break;\n            }\n        }\n        if (product == 0) return 0;\n        else {\n            if ((count % 2 != 0)) result = -1;\n            if ((count % 2 == 0)) result = 1;\n            if (result == -1) return -1;\n            else return 1;\n        }\n    }\n}\n```
['Java']