Simple parity count
Intuition\n Describe your first thoughts on how to solve this problem. \nSimply find all bits from 0th position and add corresponsing value for parity of index 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSimply find all bits from 0th position and add corresponsing value for parity of index of bit 1.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(1)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(1)\n\n# Code\n```\nclass Solution {\npublic:\n    vector<int> evenOddBit(int n) {\n        vector<int> ans = {0, 0};\n        int par = 0;\n        while(n){\n            if(par == 0){\n                if(n % 2)   ans[0]++;\n            }else{\n                if(n % 2)   ans[1]++;\n            }\n            n /= 2;\n            par = (par + 1) % 2;\n        }\n        return ans;\n    }\n};\n```
['Math', 'C++']