You are given an integer n. Your goal is to transform n into 0 using a specific set of bitwise operations. You want to find the minimum number of operations required.
The allowed operations on the binary representation of n are:
Operation 1: You can always change the rightmost (0th) bit.
Operation 2: For any bit at position i > 0, you can change it if and only if the bit at position i-1 is 1, and all bits from i-2 down to 0 are 0.
Return the minimum number of operations to transform n into 0.
Insight (Gray Code):
This problem describes the transformation from a standard binary number to its corresponding Gray code representation, and then finding the integer value of that Gray code. The number of steps to transform a number n to 0 under these rules is equivalent to the integer value of the Gray code of n. The Gray code of n can be calculated efficiently using the bitwise formula: n XOR (n >> 1).
The first line contains a single integer T, the number of test cases.
Each of the next T lines contains a single integer n.
For each test case, your function should return a single integer representing the minimum number of operations.
2
3
6
2
4
The sequence of operations to turn n into 0 exactly mirrors the steps to count from 0 to n in Gray code. The number of steps is the number itself. The sample output for n=6 is 4. Let's re-verify the GFG article's solution for n=6, which is 4. The gray code for 6 (110) is 5 (101). The gray code for 3 (011) is 2 (010). The question is subtly different from just n ^ (n >> 1).
The actual recurrence is f(n) = n - f(floor(n/2)). Let's test this.
f(6) = 6 - f(3) = 6 - (3 - f(1)) = 6 - (3 - (1 - f(0))) = 6 - (3 - 1) = 4. This matches.
f(3) = 3 - f(1) = 3 - (1 - f(0)) = 3 - 1 = 2. This matches.
The expected time complexity is O(N).
1 <= T <= 1000
0 <= n <= 10^9
Time limit: 1 sec