### Problem: Minimum Bit Flips to Convert Number (LeetCode Problem 2220)

In this problem, you are given two integers `start`

and `goal`

, and you need to determine the minimum number of bit flips required to convert `start`

to `goal`

. A bit flip changes a bit’s value from 0 to 1 or from 1 to 0.

### Approach:

**XOR Operation**: XOR (`^`

) between`start`

and`goal`

will result in a binary number where each`1`

represents a bit that is different between the two numbers.**Count the Number of**: The number of`1`

s`1`

s in the result of the XOR operation gives the number of bit flips required.

### Solution in JavaScript:

```
// Function to calculate minimum bit flips to convert start to goal
function minBitFlips(start, goal) {
// XOR the two numbers to get a result where 1s represent differing bits
let xorResult = start ^ goal;
// Count the number of 1s in the XOR result (which gives the number of bit flips)
let count = 0;
// Loop to count the number of 1s in the XOR result
while (xorResult > 0) {
count += xorResult & 1; // Check if the least significant bit is 1
xorResult >>= 1; // Right shift the result to check the next bit
}
return count; // Return the number of bit flips required
}
// Example usage:
let start = 10; // Binary: 1010
let goal = 7; // Binary: 0111
// Minimum bit flips required to convert 10 to 7
console.log(minBitFlips(start, goal)); // Output: 3
```

### Explanation:

**XOR Operation**: XOR between`start`

and`goal`

highlights all the differing bits. For example:

`start = 10 (binary: 1010)`

`goal = 7 (binary: 0111)`

`start ^ goal = 1101`

(4-bit number where 1s represent different bits).

**Counting 1s**: Each`1`

in the XOR result corresponds to a bit that needs to be flipped. By counting the number of`1`

s in the XOR result, we determine the number of bit flips needed to convert`start`

to`goal`

.

### Time Complexity:

**O(log N)**where`N`

is the maximum of`start`

or`goal`

. The time complexity is determined by the number of bits in the binary representation of the numbers.