100% ACCEPTANCE.... VERY EASY APPROACH....BEGINNER FRIENDLY...C++
Intuition : The value is multiplied by two to past equal value.....to make it simple 2 can\'t be found by OR with lesser numbers...similarly the non 2 powered v
# Intuition : The value is multiplied by two to past equal value.....to make it simple 2 can\'t be found by OR with lesser numbers...similarly the non 2 powered values like 3,5,6,7 can be found by OR within the minimal values...\nFor example, if we have 5 it can be found by 1|4 , 4 can\'t be found by any OR within the lesser numbers....any value that is equivalent to the powers of 2 can\'t be got by OR ing the minimal values among themselves..... 1,2,3 never produces 4 at anycost.....\n# Complexity\n- Time complexity:o(nlogn)\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 {\npublic:\n    int minImpossibleOR(vector<int>& nums) {\n        sort(nums.begin(),nums.end());\n        int n=nums.size();\n        int ma=1;\n        for(auto i:nums){\n            if(i==ma){\n                ma*=2;\n            }\n        }\n        return ma;\n    }\n};\n```
['Greedy', 'Sorting', 'C++']