Minimum Bit Flips to Convert Number

The problem here requires finding the number of different bits between the binary representations of start and goal. This can be done by taking the XOR of the two numbers and counting the number of bits set to 1 in the result. The reason we use XOR is that it returns a binary number where the bits in the positions where start and goal differ are set to 1.

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class Solution:
    def minBitFlips(self, start: int, goal: int) -> int:
        # Compute XOR of start and goal
        xor = start ^ goal

        # Count number of bits set to 1
        count = 0
        while xor:
            # if last bit of xor is 1, increment count
            if xor & 1:
                count += 1

            # right shift xor by 1 bit
            xor >>= 1

        return count

In the above code, we use a while loop to check each bit of the XOR result from the least significant bit to the most. We use bitwise AND (&) with 1 to check if the current bit is 1. If it is, we increment count. We then right shift XOR by one bit using >>=. This will move us to the next bit. The loop continues until all bits in xor have been checked. The final count is the number of bit flips required to transform start to goal.

Identifying Problem Isomorphism

“Minimum Number of Bit Flips to Convert Integer” can be mapped to “Hamming Distance”.

The reasoning behind this mapping is that both problems involve comparing the binary representations of two numbers.

“Hamming Distance” calculates the number of positions at which the corresponding bits are different. So, if we want to convert one number to another, the minimum number of bit flips would be exactly those positions where the two numbers have differing bits. This is equivalent to calculating the Hamming distance between the two numbers.

“Minimum Number of Bit Flips to Convert Integer” is simpler, as it doesn’t require the calculation of the exact positions of differing bits, just the count of them.

A more complex problem is “Flip Bits to Maximize Number of 1’s”. In this problem, you are given a binary string and you can flip ‘k’ bits from 0 to 1 with the goal to maximize the number of 1s in the string. This is more complex as it requires more than counting differing bits, it also needs strategic flipping to maximize the outcome.

This is exactly the same problem as https://leetcode.com/problems/hamming-distance/ but with different description.