Very Unique And Straight Forward Way🔥🫡
Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal is to make a | b equal to c with the minimum number of bit flips. For each bit
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal is to make a | b equal to c with the minimum number of bit flips. For each bit position, the values of a, b, and c are compared, and flips are counted based on whether changes are needed to match c\'s bit value.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Use the helper function getIthBit to obtain the ith bit of any integer by isolating it.\n2. Loop through all 32 bits of the integers (assuming 32-bit integers).\n3. For each bit position:\n- Check the values of the ith bits of a, b, and c.\n- If (aa | bb) == cc, the bits already match c, so no flip is required.\n- Otherwise:\n- - If both aa and bb are 1 but cc is 0, two flips are needed to set both to 0.\n- - If one of aa or bb is 1 but cc is 0, one flip is required to set that bit to 0.\n- - If both aa and bb are 0 but cc is 1, one flip is needed to set one bit to 1.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(1)$$\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n$$O(1)$$\n# Code\n```cpp []\nclass Solution {\npublic:\n    int getIthBit(int num, int i) {\n    return (num & (1 << i)) != 0;\n    }\n    int minFlips(int a, int b, int c) {\n        int ans = 0 ;\n        for(int i=31;i>=0;i--){\n            int aa = getIthBit(a,i);\n            int bb = getIthBit(b,i);\n            int cc = getIthBit(c,i);\n            if((aa|bb)==cc) continue;\n            else if(aa==1 && bb==1 && cc==0) ans+=2;\n            else if((aa==1 || bb==1) && cc==0) ans+=1;\n            else if(aa==0 && bb==0 && cc==1) ans+=1;\n        }\n        return ans;\n    }\n};\n```
['C++']