## Minimum Bit Flips to Convert Number solution leetcode

**Minimum Bit Flips to Convert Number solution leetcode** – 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: Minimum Bit Flips to Convert Number solution leetcode**

Input:start = 10, goal = 7Output:3Explanation: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: Minimum Bit Flips to Convert Number solution leetcode**

Input:start = 3, goal = 4Output:3Explanation: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: Minimum Bit Flips to Convert Number solution leetcode**

`0 <= start, goal <= 10`

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