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You are given two integers start and goal. You want to convert start to goal using the minimum number of bit flips.

A bit flip is where you change a 0 to a 1 or a 1 to a 0.

Return the minimum number of bit flips to convert start to goal.

Example 1:

Input: start = 10, goal = 7
Output: 3
Explanation: The binary representation of 10 is 1010 and the binary representation of 7 is 0111. We need to flip all the bits at positions 0, 1, and 2.

Example 2:

Input: start = 3, goal = 4
Output: 3
Explanation: The binary representation of 3 is 011 and the binary representation of 4 is 100. We need to flip all the bits at positions 0, 1, and 2.

Clarifying Questions

1. Q: Are there any constraints on the values of start and goal?
   • A: Typically, integers in such problems are within the range of 32-bit signed integers, so we can assume that ( -2^{31} \leq \text{start}, \text{goal} \leq 2^{31} - 1 ).
2. Q: Can the values of start and goal be negative?
   • A: Yes, they can be negative within the given constraints. However, we will typically deal with their unsigned binary representations.

Strategy

To determine the minimum number of bit flips required to convert start to goal, we can use the following approach:

1. XOR Operation: Perform the XOR operation between start and goal. The XOR operation results in a bit being set to 1 if the bits differ, and 0 if they are the same.

2. Count Set Bits: Count the number of 1s in the result of the XOR operation. This count will give the minimum number of bit flips needed to convert start to goal.

Steps

1. Compute xor_result = start ^ goal.
2. Count the number of 1 bits in xor_result.
3. Return the count as the result.

Code

def minBitFlips(start: int, goal: int) -> int:
    xor_result = start ^ goal
    return bin(xor_result).count('1')

# Example usage
start = 10
goal = 7
print(minBitFlips(start, goal))  # Output: 3

Time Complexity

• Time Complexity: (O(\log n)), where (n) is the maximum value between start and goal. This is because in the worst case, we check all the bits of xor_result, which is proportional to the number of bits required to represent n.
• Space Complexity: (O(1)), as we use a constant amount of extra space for variables and do not use any additional data structures.

Feel free to run the example cases to see if the solution works as described. If there are any more specific constraints or edge cases to consider, please let me know!

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