Easy to understand, simple and efficient solution
\n\n# Code\n\nclass Solution {\npublic:\n    int minFlips(int a, int b, int c) {\n        int flips = 0;\n        \n        // Iterate through the bits of a, b,
\n\n# Code\n```\nclass Solution {\npublic:\n    int minFlips(int a, int b, int c) {\n        int flips = 0;\n        \n        // Iterate through the bits of a, b, and c\n        while (a > 0 || b > 0 || c > 0) {\n            int bit_a = a & 1;  // Get the least significant bit of a\n            int bit_b = b & 1;  // Get the least significant bit of b\n            int bit_c = c & 1;  // Get the least significant bit of c\n\n            // Check if the least significant bit of c is 0\n            if (bit_c == 0) {\n                // If c is 0, both a and b should have their corresponding bits as 0\n                // Add the sum of bit_a and bit_b to flips to account for the necessary flips\n                flips += (bit_a + bit_b);\n            } else {\n                // If c is 1, at least one of a and b should have its corresponding bit as 1\n                if (bit_a == 0 && bit_b == 0) {\n                    // If both bit_a and bit_b are 0, increment flips by 1 to indicate the need for a flip\n                    flips += 1;\n                }\n            }\n\n            // Shift bits to the right\n            a >>= 1;\n            b >>= 1;\n            c >>= 1;\n        }\n\n        return flips;\n    }\n};\n\n```
['Bit Manipulation', 'C++', 'Java']