Simple to understand code bit operation!!
Here\'s the complete explanation for your solution to the problem where you need to determine the minimum number of changes required to convert the integer n to
Here\'s the complete explanation for your solution to the problem where you need to determine the minimum number of changes required to convert the integer `n` to `k` by changing bits from 1 to 0:\n\n### Intuition\nTo solve the problem, the key insight is that converting `n` to `k` is only possible if all bits set in `k` are either already set in `n` or can be ignored. This implies that the bits that are set in `k` but not in `n` would make it impossible to convert `n` to `k`.\n\n### Approach\n1. **Check Feasibility**: \n   - First, check if `k` can be derived from `n` by verifying if all bits that are set in `k` are also set in `n`. This can be done using the bitwise AND operation between the complement of `n` and `k` (i.e., `k & ~n`). If the result is not zero, it means there are bits set in `k` that cannot be set in `n`, hence return `-1`.\n\n2. **Count Changes**: \n   - If the conversion is feasible, compute the number of changes required by counting the number of bits that are set in the bitwise AND of `n` and the complement of `k` (i.e., `n & ~k`). These are the bits in `n` that need to be changed from 1 to 0 to match `k`.\n   - Use a loop to count the number of set bits in `new` (which is `n & ~k`) and shift `new` right by one bit in each iteration until all bits are processed.\n\n### Complexity\n- **Time Complexity**: \n  - The time complexity is \\(*O(log n)*\\) where \\(n\\) is the length of the binary representation of `n`. This is because the loop iterates through each bit of `new`.\n- **Space Complexity**: \n  - The space complexity is \\(*O(1)*\\) since we are using a constant amount of space regardless of the input size.\n\n### Code\n```python\nclass Solution:\n    def minChanges(self, n: int, k: int) -> int:\n        # Check if there are any bits in k that are not in n\n        if k & ~n != 0:\n            return -1\n        \n        # Count the number of bits to change from 1 to 0\n        count = 0\n        new = n & ~k\n        while new:\n            count += (new & 1)\n            new >>= 1\n        \n        return count\n```\n
['Bit Manipulation', 'Python3']