Java Solution for Minimum Flips to Make a or b Equal to c Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the above solution is to compare the bits of a, b, and c to determ
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the above solution is to compare the bits of a, b, and c to determine the number of flips required to make (a OR b) == c.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Initialize a variable flips to keep track of the total number of flips required.\n1. Iterate through the bits of a, b, and c by performing a bitwise AND operation with 1 (a & 1, b & 1, c & 1).\n1. Compare the rightmost bits of a and b with the rightmost bit of c. If (bitA | bitB) != bitC, it means flipping is required.\n1. If bitC is 0, it implies that the corresponding bit of c is 0. In this case, both bitA and bitB should be flipped to 1, so increment flips by (bitA + bitB).\n1. If bitC is 1, it implies that the corresponding bit of c is 1. In this case, at least one of bitA or bitB should be flipped to 1, so increment flips by 1.\n1. Right shift a, b, and c by 1 bit to move to the next bit.\n1. Repeat steps 2-6 until all bits have been processed.\n1. Return the final value of flips.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of the solution is O(log(max(a, b, c))) since we iterate through the bits of a, b, and c using the right shift operation.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity is O(1) since we use a constant amount of space to store the variables flips, bitA, bitB, and bitC.\n\n# Code\n```\nclass Solution {\n    public int minFlips(int a, int b, int c) {\n        int flips = 0;\n        int bitA, bitB, bitC;\n\n        while (a > 0 || b > 0 || c > 0) \n        {\n            bitA = a & 1;  // Get the rightmost bit of a\n            bitB = b & 1;  // Get the rightmost bit of b\n            bitC = c & 1;  // Get the rightmost bit of c\n\n            if ((bitA | bitB) != bitC) \n            {\n                if (bitC == 0) \n                {\n                    flips += (bitA + bitB);  // Flip both bits of a and b\n                } \n                else \n                {\n                    flips += 1;  // Flip any one bit of a or b\n                }\n            }\n\n            a >>= 1;  // Shift right to get the next bit of a\n            b >>= 1;  // Shift right to get the next bit of b\n            c >>= 1;  // Shift right to get the next bit of c\n        }\n        return flips;\n    }\n}\n```
['Java']