PROBLEM OF THE DAY || VERY EASY SOLUTION || BEST APPROACH
class Solution {\npublic:\n\n    int arraySign(vector& nums) {\n       int neg = 0; \n\t   \n     //neg will store number of negative numbers \n\t \n        for
class Solution {\npublic:\n\n    int arraySign(vector<int>& nums) {\n       int neg = 0; \n\t   \n     //neg will store number of negative numbers \n\t \n        for(int i = 0 ; i<nums.size(); i++){\n\t\t//if any number is 0 then product will be 0 \n\t\t\n            if(nums[i] == 0 )\n                return 0;\n            if(nums[i] < 0)\n                neg++;\n        }\n        \n\t\t// if negative numbers are not even the result will be negative hence -1\n\t\t\n        if(neg % 2 != 0)\n            return -1;\n        \n        return 1;\n    }\n};
['Array', 'C']