simple and easy C++ solution 😍❤️‍🔥
\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\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```cpp []\nclass Solution {\npublic:\n    int minFlips(int a, int b, int c) \n    {\n        // Initialize a variable \'ans\' to count the number of flips required.\n        int ans = 0;\n\n        // Create bitsets for a, b, and c to represent their binary form.\n        // Each bitset is 32 bits long, as integers are represented in 32 bits.\n        bitset<32>A(a), B(b), C(c);\n\n        // Iterate over each bit position from 0 to 31.\n        for(int i = 0; i < 32; i++)\n        {\n            // Check if the bitwise OR of A and B differs from C at the current bit position.\n            // (A[i] or B[i]) != C[i] checks if the combined bits of A and B do not match C.\n            if((A[i] || B[i]) != C[i])\n            {\n                // If C[i] is 0, we need to flip both A[i] and B[i] (if they are 1) to achieve 0.\n                // Thus, we add the sum of A[i] and B[i] to \'ans\'.\n                if(!C[i]) ans += A[i] + B[i];\n                // If C[i] is 1, we need to flip either A[i] or B[i] (whichever is 0) to achieve 1.\n                // Hence, we increment \'ans\' by 1.\n                else ans += 1;\n            }\n        }\n\n        // Return the total number of flips required to convert A and B into C.\n        return ans;\n    }\n};\n```\n\n\n# if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n###### Let\'s Connect on LinkedIn: www.linkedin.com/in/shishirrsiam\n###### Let\'s Connect on Facebook: www.fb.com/shishirrsiam
['Bit Manipulation', 'C++']