> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Minimum Bit Flips to Convert Number

> Tested Python solution for LeetCode 2220 with 24 pytest cases. Generate a practice environment with lcpy.

LeetCode 2220, [Easy](/catalog/easy). Topics: [Bit Manipulation](/catalog/topics/bit-manipulation). [View on LeetCode](https://leetcode.com/problems/minimum-bit-flips-to-convert-number/description/).

Generate this problem as a practice environment: tested reference solution, 24 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 2220   # by problem number
lcpy gen -s minimum_bit_flips_to_convert_number   # by problem name
```

## Problem

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`.

### Examples

```
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.
```

```
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 \<= 10^9

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/minimum_bit_flips_to_convert_number/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/minimum_bit_flips_to_convert_number/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class Solution:
    # Time: O(log(max(start, goal)))
    # Space: O(1)
    def min_bit_flips(self, start: int, goal: int) -> int:
        return (start ^ goal).bit_count()
```

## Complexity

| Time | Space |
| - | - |
| O(log(max(start, goal))) | O(1) |

## Tags

[NeetCode All](/catalog/neetcode).


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