Python 3   ||   9 lines,  divmod   ||   T/S: 92% / 99%
\n    class Solution:\n        def minFlips(self, a: int, b: int, c: int) -> int:\n        \n            ans = 0\n\n            while not a == b == c == 0:\n   
```\n    class Solution:\n        def minFlips(self, a: int, b: int, c: int) -> int:\n        \n            ans = 0\n\n            while not a == b == c == 0:\n                a, A = divmod(a,2)\n                b, B = divmod(b,2)\n                c, C = divmod(c,2)\n\n                if   A == B == 1 and C == 0 : ans+= 2\n                elif A == B == 0 and C == 1 : ans+= 1\n                elif A != B      and C == 0 : ans+= 1\n                \n            return ans\n```\n[https://leetcode.com/problems/minimum-flips-to-make-a-or-b-equal-to-c/submissions/692614680/](https://leetcode.com/problems/minimum-flips-to-make-a-or-b-equal-to-c/submissions/692614680/)\n\n\n\n\nI could be wrong, but I think that time complexity is *O*(*N*) and space complexity is *O*(1), in which *N* ~`max(a,b,c)`.
['Python3']