2220. Minimum Bit Flips to Convert Number

[mah-shamim]

Topics: Bit Manipulation

A bit flip of a number x is choosing a bit in the binary representation of x and flipping it from either 0 to 1 or 1 to 0.

• For example, for x = 7, the binary representation is 111 and we may choose any bit (including any leading zeros not shown) and flip it. We can flip the first bit from the right to get 110, flip the second bit from the right to get 101, flip the fifth bit from the right (a leading zero) to get 10111, etc.

Given two integers start and goal, 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 and 7 are 1010 and 0111 respectively. We can convert 10 to 7 in 3 steps:
  • Flip the first bit from the right: 1010 -> 1011.
  • Flip the third bit from the right: 1011 -> 1111.
  • Flip the fourth bit from the right: 1111 -> 0111.
    It can be shown we cannot convert 10 to 7 in less than 3 steps. Hence, we return 3.

Example 2:

• Input: start = 3, goal = 4
• Output: 3
• Explanation: The binary representation of 3 and 4 are 011 and 100 respectively. We can convert 3 to 4 in 3 steps:
  • Flip the first bit from the right: 011 -> 010.
  • Flip the second bit from the right: 010 -> 000.
  • Flip the third bit from the right: 000 -> 100.
    It can be shown we cannot convert 3 to 4 in less than 3 steps. Hence, we return 3.

Constraints:

• 0 <= start, goal <= 109

Hint:

1. If the value of a bit in start and goal differ, then we need to flip that bit.
2. Consider using the XOR operation to determine which bits need a bit flip.

[basharul-siddike]

We need to determine how many bit positions differ between start and goal. This can be easily achieved using the XOR operation (^), which returns a 1 for each bit position where the two numbers differ.

Steps:

1. Perform the XOR operation between start and goal. The result will be a number that has 1s in all the positions where start and goal differ.
2. Count how many 1s are present in the binary representation of the result (i.e., the Hamming distance).
3. The number of 1s will give us the minimum number of bit flips needed.

Let's implement this solution in PHP: 2220. Minimum Bit Flips to Convert Number

<?php
/**
 * @param Integer $start
 * @param Integer $goal
 * @return Integer
 */
function minBitFlips($start, $goal) {
    // Step 1: XOR start and goal to find the differing bits
    $xor = $start ^ $goal;
    
    // Step 2: Count the number of 1's in the binary representation of xor
    $bitFlips = 0;
    while ($xor > 0) {
        // Increment count if the last bit is 1
        $bitFlips += $xor & 1;
        // Shift xor to the right by 1 bit
        $xor >>= 1;
    }
    
    return $bitFlips;
}

// Test cases
echo minBitFlips(10, 7);  // Output: 3
echo "\n";
echo minBitFlips(3, 4);   // Output: 3
?>

Explanation:

1. The ^ (XOR) operation compares each bit of start and goal. If the bits are different, the corresponding bit in the result will be 1.
2. We then count the number of 1s in the result, which gives the number of differing bits, i.e., the number of bit flips required.
3. The & 1 operation checks if the last bit is 1, and >>= 1 shifts the number right to process the next bit.

Time Complexity:

• The time complexity is (O(\log N)), where (N) is the larger of start or goal, because we're checking each bit of the number. In the worst case, we will loop through all bits of a 32-bit integer (since PHP 5.6 works with 32-bit or 64-bit integers depending on the system).

Output:

• For start = 10 and goal = 7, the output is 3.
• For start = 3 and goal = 4, the output is 3.

[mah-shamim]

Thanks to solve this problem

[basharul-siddike]

Thanks