Minimizing Bit Flips | Detailed Explanation with Change Requirements for Each Bit Combination
Intuition\n Describe your first thoughts on how to solve this problem. \nTo solve this problem, we need to determine how many flips are required to make the bit
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo solve this problem, we need to determine how many flips are required to make the bitwise OR of two integers\xA0`a`\xA0and\xA0`b` equal to a third integer\xA0`c`. The idea is to process the bits from the least significant to the most significant and decide how many changes are needed based on the current bits of\xA0`a`,\xA0`b`, and\xA0`c`.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Iterate through the bits: We use a loop to process each bit of the integers a, b, and c starting from the least significant bit (rightmost).\n2. Check conditions for flipping:\n    - If the bit in c is 1 and the corresponding bits in a and b are both 0, we need to flip one of a or b to 1.\n    - If the bit in c is 0 and both bits in a and b are 1, we need to flip both bits to 0.\n3. Count flips: Increase the count of flips based on the above conditions.\nRight shift: Move to the next more significant bit by right-shifting a, b, and c.\n4. Terminate: The loop runs until all bits in a, b, and c have been processed.\n\n# Complexity\n- Time complexity:  $$O(30) \\equiv O(1) \\space \\space \\because (10^9 \\equiv 2^{30})$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n    def minFlips(self, a: int, b: int, c: int) -> int:\n        # a.   b.  |  c.  | no of changes needed\n        # 0.   0.     0.           0\n        # 0.   1.     1.           0\n        # 1.   0.     1.           0\n\n        # 0.   0.     1.           1   (set either a or b to 1)\n        # 0.   1.     0.           1\n        # 1.   0.     0.           1\n\n        # 1.   1.     0.           2   (both needed to be set to 0)\n\n        res = 0\n        while c or a or b:\n            curr_c = c & 1\n            curr_a = a & 1\n            curr_b = b & 1\n            \n            # checking for the last four conditions\n            if (curr_c) != (curr_a or curr_b):\n\n                res += 1\n                # checking for the last condition (add extra 1)\n                if not curr_c and (curr_a and curr_b):\n                    res += 1\n                    \n            \n            # augmented assigments\n            c >>= 1\n            a >>= 1\n            b >>= 1\n        \n        return res\n\n```
['Python3']