• Home
• com.github.javadev
• leetcode-in-kotlin
• 1.21
• source code
• readme.md

×

Project download

Many resources are needed to download a project. Please understand that we have to compensate our server costs. Thank you in advance.
Project price only 1 $

You can buy this project and download/modify it how often you want.

g0801_0900.s0868_binary_gap.readme.md Maven / Gradle / Ivy

868\. Binary Gap

Easy

Given a positive integer `n`, find and return _the **longest distance** between any two **adjacent**_ `1`_'s in the binary representation of_ `n`_. If there are no two adjacent_ `1`_'s, return_ `0`_._

Two `1`'s are **adjacent** if there are only `0`'s separating them (possibly no `0`'s). The **distance** between two `1`'s is the absolute difference between their bit positions. For example, the two `1`'s in `"1001"` have a distance of 3.

**Example 1:**

**Input:** n = 22

**Output:** 2

**Explanation:** 22 in binary is "10110".

The first adjacent pair of 1's is "10110" with a distance of 2.

The second adjacent pair of 1's is "10110" with a distance of 1.

The answer is the largest of these two distances, which is 2.

Note that "10110" is not a valid pair since there is a 1 separating the two 1's underlined.

**Example 2:**

**Input:** n = 8

**Output:** 0

**Explanation:** 8 in binary is "1000". There are not any adjacent pairs of 1's in the binary representation of 8, so we return 0.

**Example 3:**

**Input:** n = 5

**Output:** 2

**Explanation:** 5 in binary is "101".

**Constraints:**

*   1 <= n <= 109