EASY SOLUTION C++ | Quick To Understand
Intuition\nThink about how the numbers are related and how can you check exactly which number will cause no or calculation\n\n# Approach\nThe numbers that have 
# Intuition\nThink about how the numbers are related and how can you check exactly which number will cause no or calculation\n\n# Approach\nThe numbers that have an increase in bit eg: 2-10 4-100 , 2 has 2 and 4 has 3 so the max number that causes break is our result\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int minImpossibleOR(vector<int>& nums) {\n        int v=0;\n        sort(nums.begin(),nums.end());\n        for(int i=0;i<nums.size();i++)\n        {\n            if(nums[i]==pow(2,v))v++;\n        }\n        return pow(2,v);\n    }\n};\n```
['Bit Manipulation', 'Brainteaser', 'C++']