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Let\'s break down the solution for the problem at hand, covering the intuition, approach, complexity analysis, and the provided code.\n\n# Intuition\nThe proble
Let\'s break down the solution for the problem at hand, covering the intuition, approach, complexity analysis, and the provided code.\n\n# Intuition\nThe problem involves finding the minimum number of bit changes required to transform an integer `n` into another integer `k`. The key is to compare the bits of `n` and `k` from the least significant bit to the most significant bit and count the changes needed.\n\n# Approach\n1. Initialize a counter `ans` to keep track of the number of bit changes.\n2. Iterate while `n` is not zero:\n    - Check the least significant bit (LSB) of `n` and `k`.\n    - If the LSB of `n` is 1 and the LSB of `k` is 0, increment the `ans` counter (as a change from 1 to 0 is needed).\n    - If the LSB of `n` is 0 and the LSB of `k` is 1, return -1 (since changing a 0 bit to a 1 bit is not allowed in this problem).\n3. Right shift both `n` and `k` to process the next bit.\n4. If after processing all bits of `n`, `k` still has remaining bits with value 1, return -1 (as it\'s impossible to match `k`).\n5. Return the `ans` counter as the result.\n\n# Complexity\n- **Time complexity**: $$O(log n)$$, where $$log n$$ is the number of bits in the binary representation of `n`. This is because each iteration processes one bit.\n- **Space complexity**: $$O(1)$$, as only a few variables are used for computation and no additional data structures are required.\n\n# Code\n```cpp\nclass Solution {\npublic:\n    int minChanges(int n, int k) {\n        int ans = 0;\n        while (n) {\n            if (n % 2) {  // If the current bit of n is 1\n                if (k % 2 == 0) ans++;  // If the current bit of k is 0, we need a change\n            } else {  // If the current bit of n is 0\n                if (k % 2) return -1;  // If the current bit of k is 1, it\'s impossible to change 0 to 1\n            }\n            n /= 2;  // Right shift n to process the next bit\n            k /= 2;  // Right shift k to process the next bit\n        }\n        if (k > n) return -1;  // If k still has bits left, it\'s impossible to match\n        return ans;\n    }\n};\n```\n\n### Explanation of the Code\n1. **Initialization**: `ans` is initialized to 0 to count the required changes.\n2. **Main Loop**:\n    - The loop continues while `n` is not zero.\n    - For each bit, if the bit of `n` is 1 and the bit of `k` is 0, increment `ans`.\n    - If the bit of `n` is 0 and the bit of `k` is 1, return -1 because we cannot change a 0 to a 1.\n    - Both `n` and `k` are right-shifted to process the next bit.\n3. **Post Loop Check**: \n    - If `k` still has bits left after `n` is exhausted, return -1 as it\'s impossible to match `k`.\n4. **Return Result**: The count of changes (`ans`) is returned.
['Bit Manipulation', 'C++']