Python/C++ 1 Liner | Bit Manipulation
Intuition\n Describe your first thoughts on how to solve this problem. \nMy first approach was to compare the bits one by one like most solutions show. However,
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nMy first approach was to compare the bits one by one like most solutions show. However, I realized that this can be done faster and in one line if you use bitwise operators. I\'ll be referencing the while loop approach a lot, but I haven\'t included it. Feel free to check out the other great solutions out there.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nIn the while loop approach, we compare each bit in `a`, `b`, and `c`. If a bit is a `1` in `c` and is a `0` in both `a` and `b`, then that counts as one flip. If a bit is `0` in `c`, then that counts as one flip if that bit is a `1` in `a` and another flip if that bit is a `1` in `b`.\n\nIn other words, we can break this down into three operations:\n1. Count how many bits in `a | b` need to be flipped to a `1`.\n2. Count how many bits in `a` need to be flipped to a `0`.\n3. Count how many bits in `b` need to be flipped to a `0`.\n\n### Step 1\nStep 1 is calculated with\n```\n(((a | b) & c) ^ c).bit_count()\n```\nLet\'s break this down.\n\nFirst, we calculate `a | b`. Then, we use `c` as a bitmask to extract only the bits in `a | b` that are set to `1` in `c`. Finally, we use an xor to calculate how many of the set bits in `c` are `0` in `a | b`. This works because the bits that are `1`s in both `c` and `a | b` will become `0`s, while the bits that are `1`s in `c` and `0`s in `a | b` will become `1`s. And, since we used `c` as a bitmask, we don\'t have to worry about bit that are `1`s in `a | b` but `0`s in `c`. Finally, we count the number of these set bits to give us the total number of bits in `a | b` that need to be flipped from `0`s to `1`s.\n\n### Step 2 / 3\nSteps 2 and 3 are calculated with  \n```\n((c & a) ^ a).bit_count() + ((c & b) ^ b).bit_count()\n```\nLet\'s break this down\n\nWe follow a similar approach to step 1. First, we use a bitmask. The main difference from step one is that now, we use the bitmask on `c`. Specifically, we extract all the bits in `c` that are `1`s in `a` (or `1`s in `b` for step 3). Then, we xor with `a` to calculate how many of the set bits in `a` (or `b` for step 3) are `0`s in `c`. Finally, we count how many of those bits there are to give us the total number of bits in `a` (or `b` for step 3) that need to be flipped from `1`s to `0`s.\n\n\n# Complexity\n- Time complexity: $$O(1)$$\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\n```Python []\nclass Solution:\n    def minFlips(self, a: int, b: int, c: int) -> int:\n        return ((c & a) ^ a).bit_count() + ((c & b) ^ b).bit_count() + (((a | b) & c) ^ c).bit_count()\n```\n```C++ []\nclass Solution {\npublic:\n    int minFlips(int a, int b, int c) {\n        return __builtin_popcount((c & a) ^ a) + __builtin_popcount((c & b) ^ b) + __builtin_popcount(((a | b) & c) ^ c);\n    }\n};\n```\n
['C++', 'Python3']