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Intuition\nThe problem requires finding the minimum number of bit flips in the binary representations of two numbers, \'a\' and \'b\', such that their bitwise O
# Intuition\nThe problem requires finding the minimum number of bit flips in the binary representations of two numbers, \'a\' and \'b\', such that their bitwise OR is equal to another number \'c\'. To solve this problem, we can iterate through the binary representations of \'a\', \'b\', and \'c\' simultaneously and check each bit. If the corresponding bit in \'c\' is set to 1, we need to ensure that at least one of the bits in \'a\' or \'b\' is also set to 1. If the bit in \'c\' is 0, we need to ensure that both bits in \'a\' and \'b\' are set to 0.\n\n\n# Approach\n1.\tConvert \'a\', \'b\', and \'c\' to their binary representations using the \'format\' function.\n2.\tDetermine the maximum length among the binary representations of \'a\', \'b\', and \'c\'.\n3.\tPad the shorter binary representations with leading zeros to ensure all three have the same length.\n4.\tInitialize a variable \'res\' to keep track of the minimum number of flips required.\n5.\tIterate through each bit position from left to right.\n\u2022\tIf the bit in \'c\' is 1 and both bits in \'a\' and \'b\' are 0, increment \'res\' by 1.\n\u2022\tIf the bit in \'c\' is 0 and either bit in \'a\' or \'b\' is 1, increment \'res\' by 1 for each set bit.\n6.\tReturn the final value of \'res\', which represents the minimum number of flips required to achieve (a OR b == c).\n\n\n# Complexity\n- Time complexity:\nThe time complexity of this approach is O(log(max(a, b, c))), as we need to iterate through the binary representations of \'a\', \'b\', and \'c\'.\n\n\n- Space complexity:\nThe space complexity is O(log(max(a, b, c))), as we need to store the binary representations of \'a\', \'b\', and \'c\' in separate variables.\n\n\n\n\n\n# Code\n```\nclass Solution:\n    def minFlips(self, a: int, b: int, c: int) -> int:\n        abin=format(a,\'b\')\n        bbin=format(b,\'b\')\n        cbin=format(c,\'b\')\n        x = max(max(len(abin),len(bbin)),len(cbin))\n        for i in range(x-len(abin)):\n            abin="0"+abin\n        for i in range(x-len(bbin)):\n            bbin="0"+bbin\n        for i in range(x-len(cbin)):\n            cbin="0"+cbin\n        res=0\n        for i in range(x):\n            if cbin[i]==\'1\':\n                if abin[i]==\'0\' and bbin[i]==\'0\':\n                    res+=1\n            else:\n                if abin[i]==\'1\' :\n                    res+=1\n                if bbin[i]==\'1\' :\n                    res+=1\n        return res\n            \n```
['Python3']