# [LeetCode]#2220. Minimum Bit Flips to Convert Number

2 min readNov 28, 2022

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**Environment: Python 3.8**

**Key technique: XOR, count**

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.

**Analysis:**

- Use XOR to get output.
- Sum all ‘1’

**Solution:**

`class Solution:`

def minBitFlips(self, start, goal):

return bin(start ^ goal).count('1')

**Submissions:**

**Reference:**