• Home
• com.github.javadev
• leetcode-in-java
• 1.27
• 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.

g2701_2800.s2709_greatest_common_divisor_traversal.readme.md Maven / Gradle / Ivy

2709\. Greatest Common Divisor Traversal

Hard

You are given a **0-indexed** integer array `nums`, and you are allowed to **traverse** between its indices. You can traverse between index `i` and index `j`, `i != j`, if and only if `gcd(nums[i], nums[j]) > 1`, where `gcd` is the **greatest common divisor**.

Your task is to determine if for **every pair** of indices `i` and `j` in nums, where `i < j`, there exists a **sequence of traversals** that can take us from `i` to `j`.

Return `true` _if it is possible to traverse between all such pairs of indices,_ _or_ `false` _otherwise._

**Example 1:**

**Input:** nums = [2,3,6]

**Output:** true

**Explanation:** In this example, there are 3 possible pairs of indices: (0, 1), (0, 2), and (1, 2). To go from index 0 to index 1, we can use the sequence of traversals 0 -> 2 -> 1, where we move from index 0 to index 2 because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 > 1, and then move from index 2 to index 1 because gcd(nums[2], nums[1]) = gcd(6, 3) = 3 > 1. To go from index 0 to index 2, we can just go directly because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 > 1. Likewise, to go from index 1 to index 2, we can just go directly because gcd(nums[1], nums[2]) = gcd(3, 6) = 3 > 1.

**Example 2:**

**Input:** nums = [3,9,5]

**Output:** false

**Explanation:** No sequence of traversals can take us from index 0 to index 2 in this example. So, we return false.

**Example 3:**

**Input:** nums = [4,3,12,8]

**Output:** true

**Explanation:** There are 6 possible pairs of indices to traverse between: (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), and (2, 3). A valid sequence of traversals exists for each pair, so we return true.

**Constraints:**

*   1 <= nums.length <= 105
*   1 <= nums[i] <= 105