Minimum One Bit Operations to Make Integers Zero - Solution & Explanation

The idea is to use a recursive function to determine the minimum number of operations needed to reduce the given number n to 0. Consider each bit change step as a recursive call, effectively exploring the tree of possibilities until we reach 0.

We need to account for the operation rules defined in the problem, particularly handling the change in the i-th bit when needed.

This function computes the recursive solution. It checks the highest set bit and recurses by flipping the bit using XOR and adjusting the metric accordingly. The base case is when n is 0.

This approach removes recursion and implements the bit operations iteratively. The key step involves simulating the recursive logic using a while loop and maintaining a transformation count.

The iterative loop continuously XORs res with current num and shifts num right, mimicking recursive pattern iteratively.

This problem essentially asks for the inverse transformation of Gray code at position n, i.e., constructing the original number from the Gray code.

Let's first review how to convert binary code to binary Gray code. The rule is to keep the most significant bit of the binary code as the most significant bit of the Gray code, while the second most significant bit of the Gray code is obtained by XORing the most significant bit and the second most significant bit of the binary code. The remaining bits of the Gray code are computed similarly to the second most significant bit.

Suppose a binary number is represented as B_{n-1}B_{n-2}...B_2B_1B_0, and its Gray code representation is G_{n-1}G_{n-2}...G_2G_1G_0. The most significant bit is kept, so G_{n-1} = B_{n-1}; and for other bits G_i = B_{i+1} \oplus B_{i}, where i=0,1,2..,n-2.

So what is the inverse transformation from Gray code to binary code?

We can observe that the most significant bit of the Gray code is kept, so B_{n-1} = G_{n-1}; and B_{n-2} = G_{n-2} \oplus B_{n-1} = G_{n-2} \oplus G_{n-1}; and for other bits B_i = G_{i} \oplus G_{i+1} cdots \oplus G_{n-1}, where i=0,1,2..,n-2. Therefore, we can use the following function rev(x) to obtain its binary code:

int rev(int x) {
    int n = 0;
    for (; x != 0; x >>= 1) {
        n ^= x;
    }
    return n;
}

The time complexity is O(log n), where n is the integer given in the problem. The space complexity is O(1).