simple explanation for noobs like me
in this problem i simply thought that we have to select some elements and after some interger multipication my result should be 1  now let say we have selected 
in this problem i simply thought that we have to select some elements and after some interger multipication my result should be 1  now let say we have selected some elements\n\n\n\n\n\nnow we have 2 cases\ncase1: -hcf of selected elements==1\n\ncase2:- hcf of selected elements!=1\n\nin case 2 we can take some intezer common let set we select \n(a,b,c,d)\nafter taking hcf common let say h\nh(a1,b1,c1,d1)   here h !=1(case2) \'\nwe can easily say that whatever multiplication(with int) we make on a1,b1,c1,d1 we can\'t make it a fraction (as the result  is always intezer) so the expression we can,t make it 1 as h>1 \n\nlow let say we are given an arr\narr={ 6, 8, 12, 14, 3,  9, 15}\'\n\'\nnow let say we select element 6 8 and 12 we can easily see hcf is 2\nso we can write it as  2( 3 , 4,  6) now whatever operation we make on 3,4,6  we can,t make it 1/2 for sure so we can\'t make the expression as 1\n\n\n\n\nbut in case 1 we can make it 1 i don\'t know the reason for this !!\n\n\n```\nclass Solution {\npublic:\n    \n    int hcf(int a,int b)\n    {\n        if(a==0)return b;\n        \n        if(a>b)return hcf(b,a);\n        \n        return hcf(b%a,a);   \n    }\n    \n    bool isGoodArray(vector<int>& nums) {\n       \n        bool ans=false;\n        \n        if(nums.size()==1)\n        {\n            return nums[0]==1;\n        }\n        \n        \n        \n        \n        int h=hcf(nums[0],nums[1]);\n        \n        if(h==1)return true;\n        \n        for(int i=2;i<nums.size();i++)\n        {\n            h=hcf(h,nums[i]);\n            \n            if(h==1)\n            {\n                ans=true;\n                break;\n            }\n            \n        }\n        \n        \n        return ans;\n        \n        \n    }\n};\n```
['C']