leetcode-problemset/leetcode-cn/originData/greatest-common-divisor-traversal.json

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            "title": "Greatest Common Divisor Traversal",
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            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>, and you are allowed to <strong>traverse</strong> between its indices. You can traverse between index <code>i</code> and index <code>j</code>, <code>i != j</code>, if and only if <code>gcd(nums[i], nums[j]) &gt; 1</code>, where <code>gcd</code> is the <strong>greatest common divisor</strong>.</p>\n\n<p>Your task is to determine if for <strong>every pair</strong> of indices <code>i</code> and <code>j</code> in nums, where <code>i &lt; j</code>, there exists a <strong>sequence of traversals</strong> that can take us from <code>i</code> to <code>j</code>.</p>\n\n<p>Return <code>true</code><em> if it is possible to traverse between all such pairs of indices,</em><em> or </em><code>false</code><em> otherwise.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,3,6]\n<strong>Output:</strong> true\n<strong>Explanation:</strong> In this example, there are 3 possible pairs of indices: (0, 1), (0, 2), and (1, 2).\nTo go from index 0 to index 1, we can use the sequence of traversals 0 -&gt; 2 -&gt; 1, where we move from index 0 to index 2 because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 &gt; 1, and then move from index 2 to index 1 because gcd(nums[2], nums[1]) = gcd(6, 3) = 3 &gt; 1.\nTo go from index 0 to index 2, we can just go directly because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 &gt; 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 &gt; 1.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,9,5]\n<strong>Output:</strong> false\n<strong>Explanation:</strong> No sequence of traversals can take us from index 0 to index 2 in this example. So, we return false.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [4,3,12,8]\n<strong>Output:</strong> true\n<strong>Explanation:</strong> 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.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
            "translatedTitle": "最大公约数遍历",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;，你可以在一些下标之间遍历。对于两个下标&nbsp;<code>i</code>&nbsp;和&nbsp;<code>j</code>（<code>i != j</code>），当且仅当&nbsp;<code>gcd(nums[i], nums[j]) &gt; 1</code>&nbsp;时，我们可以在两个下标之间通行，其中&nbsp;<code>gcd</code>&nbsp;是两个数的 <strong>最大公约数</strong>&nbsp;。</p>\n\n<p>你需要判断 <code>nums</code>&nbsp;数组中&nbsp;<strong>任意&nbsp;</strong>两个满足 <code>i &lt; j</code>&nbsp;的下标&nbsp;<code>i</code>&nbsp;和&nbsp;<code>j</code> ，是否存在若干次通行可以从 <code>i</code>&nbsp;遍历到 <code>j</code>&nbsp;。</p>\n\n<p>如果任意满足条件的下标对都可以遍历，那么返回 <code>true</code>&nbsp;，否则返回 <code>false</code>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [2,3,6]\n<b>输出：</b>true\n<b>解释：</b>这个例子中，总共有 3 个下标对：(0, 1) ，(0, 2) 和 (1, 2) 。\n从下标 0 到下标 1 ，我们可以遍历 0 -&gt; 2 -&gt; 1 ，我们可以从下标 0 到 2 是因为 gcd(nums[0], nums[2]) = gcd(2, 6) = 2 &gt; 1 ，从下标 2 到 1 是因为 gcd(nums[2], nums[1]) = gcd(6, 3) = 3 &gt; 1 。\n从下标 0 到下标 2 ，我们可以直接遍历，因为 gcd(nums[0], nums[2]) = gcd(2, 6) = 2 &gt; 1 。同理，我们也可以从下标 1 到 2 因为 gcd(nums[1], nums[2]) = gcd(3, 6) = 3 &gt; 1 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [3,9,5]\n<b>输出：</b>false\n<b>解释：</b>我们没法从下标 0 到 2 ，所以返回 false 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [4,3,12,8]\n<b>输出：</b>true\n<b>解释：</b>总共有 6 个下标对：(0, 1) ，(0, 2) ，(0, 3) ，(1, 2) ，(1, 3) 和 (2, 3) 。所有下标对之间都存在可行的遍历，所以返回 true 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
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