Easy to understand | Beats 100 %
Approach\n\n1. Initialize a Counter:\n   - Start with a counter flips set to 0. This will keep track of the number of bit flips needed.\n\n2. Iterate Through Ea
# Approach\n\n1. **Initialize a Counter**:\n   - Start with a counter `flips` set to 0. This will keep track of the number of bit flips needed.\n\n2. **Iterate Through Each Bit Position**:\n   - Use a loop to iterate through each bit position of the integers \\(a\\), \\(b\\), and \\(c\\). Continue the loop until all bits of \\(a\\), \\(b\\), and \\(c\\) have been processed.\n\n3. **Extract the Last Bits**:\n   - For each iteration, extract the last bit (least significant bit) of \\(a\\), \\(b\\), and \\(c\\) using the bitwise AND operation with 1 (`a & 1`, `b & 1`, `c & 1`).\n\n4. **Compare the Bits**:\n   - Compare the extracted bits of \\(a\\) and \\(b\\) with the corresponding bit of \\(c\\):\n     - If the bitwise OR of the last bits of \\(a\\) and \\(b\\) (`bitA | bitB`) does not match the last bit of \\(c\\) (`bitC`), then a flip is needed.\n     - If `bitC` is 1, it means at least one of `bitA` or `bitB` must be 1. If neither is 1, increment the `flips` counter by 1.\n     - If `bitC` is 0, both `bitA` and `bitB` must be 0. If either `bitA` or `bitB` is 1, increment the `flips` counter accordingly.\n\n5. **Right Shift the Bits**:\n   - Right shift \\(a\\), \\(b\\), and \\(c\\) by 1 bit to process the next bit position in the next iteration.\n\n6. **Return the Result**:\n   - After processing all bit positions, return the `flips` counter as the result.\n\nThis approach ensures that you efficiently determine the minimum number of bit flips required by examining each bit position individually and making the necessary adjustments based on the comparison with the target bit in \\(c\\).\n# Complexity\n- Time complexity:O(Log(Max(A,B,C)))\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```java []\nclass Solution {\n    public int minFlips(int a, int b, int c) {\n        int flips = 0; \n        // Iterate through each bit position\n        while (a > 0 || b > 0 || c > 0) {\n            // Get the last bits of a, b, and c\n            int bitA = a & 1;\n            int bitB = b & 1;\n            int bitC = c & 1; \n            // If the OR of bitA and bitB doesn\'t match bitC\n            if ((bitA | bitB) != bitC) {\n                if (bitC == 1) {\n                    // If bitC is 1, we need to flip one of bitA or bitB to 1\n                    flips += 1;\n                } else {\n                    // If bitC is 0, we need to flip both bits if both are 1\n                    flips += (bitA == 1 ? 1 : 0) + (bitB == 1 ? 1 : 0);\n                }\n            }\n\n            // Right shift to check the next bits\n            a >>= 1;\n            b >>= 1;\n            c >>= 1;\n        }\n        return flips;\n    }\n    }\n```
['C++', 'Java']