leetcode-problemset/leetcode-cn/originData/path-existence-queries-in-a-graph-ii.json

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        "question": {
            "questionId": "3852",
            "questionFrontendId": "3534",
            "categoryTitle": "Algorithms",
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            "title": "Path Existence Queries in a Graph II",
            "titleSlug": "path-existence-queries-in-a-graph-ii",
            "content": "<p>You are given an integer <code>n</code> representing the number of nodes in a graph, labeled from 0 to <code>n - 1</code>.</p>\n\n<p>You are also given an integer array <code>nums</code> of length <code>n</code> and an integer <code>maxDiff</code>.</p>\n\n<p>An <strong>undirected </strong>edge exists between nodes <code>i</code> and <code>j</code> if the <strong>absolute</strong> difference between <code>nums[i]</code> and <code>nums[j]</code> is <strong>at most</strong> <code>maxDiff</code> (i.e., <code>|nums[i] - nums[j]| &lt;= maxDiff</code>).</p>\n\n<p>You are also given a 2D integer array <code>queries</code>. For each <code>queries[i] = [u<sub>i</sub>, v<sub>i</sub>]</code>, find the <strong>minimum</strong> distance between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code><sub>.</sub> If no path exists between the two nodes, return -1 for that query.</p>\n\n<p>Return an array <code>answer</code>, where <code>answer[i]</code> is the result of the <code>i<sup>th</sup></code> query.</p>\n\n<p><strong>Note:</strong> The edges between the nodes are unweighted.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">n = 5, nums = [1,8,3,4,2], maxDiff = 3, queries = [[0,3],[2,4]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[1,1]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The resulting graph is:</p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2025/03/25/4149example1drawio.png\" style=\"width: 281px; height: 161px;\" /></p>\n\n<table>\n\t<tbody>\n\t\t<tr>\n\t\t\t<th>Query</th>\n\t\t\t<th>Shortest Path</th>\n\t\t\t<th>Minimum Distance</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[0, 3]</td>\n\t\t\t<td>0 &rarr; 3</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[2, 4]</td>\n\t\t\t<td>2 &rarr; 4</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t</tbody>\n</table>\n\n<p>Thus, the output is <code>[1, 1]</code>.</p>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">n = 5, nums = [5,3,1,9,10], maxDiff = 2, queries = [[0,1],[0,2],[2,3],[4,3]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[1,2,-1,1]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The resulting graph is:</p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2025/03/25/4149example2drawio.png\" style=\"width: 281px; height: 121px;\" /></p>\n</div>\n\n<table>\n\t<tbody>\n\t\t<tr>\n\t\t\t<th>Query</th>\n\t\t\t<th>Shortest Path</th>\n\t\t\t<th>Minimum Distance</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[0, 1]</td>\n\t\t\t<td>0 &rarr; 1</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[0, 2]</td>\n\t\t\t<td>0 &rarr; 1 &rarr; 2</td>\n\t\t\t<td>2</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[2, 3]</td>\n\t\t\t<td>None</td>\n\t\t\t<td>-1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[4, 3]</td>\n\t\t\t<td>3 &rarr; 4</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t</tbody>\n</table>\n\n<p>Thus, the output is <code>[1, 2, -1, 1]</code>.</p>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">n = 3, nums = [3,6,1], maxDiff = 1, queries = [[0,0],[0,1],[1,2]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[0,-1,-1]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>There are no edges between any two nodes because:</p>\n\n<ul>\n\t<li>Nodes 0 and 1: <code>|nums[0] - nums[1]| = |3 - 6| = 3 &gt; 1</code></li>\n\t<li>Nodes 0 and 2: <code>|nums[0] - nums[2]| = |3 - 1| = 2 &gt; 1</code></li>\n\t<li>Nodes 1 and 2: <code>|nums[1] - nums[2]| = |6 - 1| = 5 &gt; 1</code></li>\n</ul>\n\n<p>Thus, no node can reach any other node, and the output is <code>[0, -1, -1]</code>.</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= maxDiff &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>queries[i] == [u<sub>i</sub>, v<sub>i</sub>]</code></li>\n\t<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt; n</code></li>\n</ul>\n",
            "translatedTitle": "针对图的路径存在性查询 II",
            "translatedContent": "<p>给你一个整数 <code>n</code>，表示图中的节点数量，这些节点按从 <code>0</code> 到 <code>n - 1</code>&nbsp;编号。</p>\n\n<p>同时给你一个长度为 <code>n</code> 的整数数组 <code>nums</code>，以及一个整数 <code>maxDiff</code>。</p>\n\n<p>如果满足 <code>|nums[i] - nums[j]| &lt;= maxDiff</code>（即 <code>nums[i]</code> 和 <code>nums[j]</code> 的&nbsp;<strong>绝对差&nbsp;</strong>至多为 <code>maxDiff</code>），则节点 <code>i</code> 和节点 <code>j</code> 之间存在一条&nbsp;<strong>无向边&nbsp;</strong>。</p>\n\n<p>此外，给你一个二维整数数组 <code>queries</code>。对于每个 <code>queries[i] = [u<sub>i</sub>, v<sub>i</sub>]</code>，找到节点 <code>u<sub>i</sub></code> 和节点 <code>v<sub>i</sub></code> 之间的&nbsp;<strong>最短距离&nbsp;</strong>。如果两节点之间不存在路径，则返回 -1。</p>\n\n<p>返回一个数组 <code>answer</code>，其中 <code>answer[i]</code> 是第 <code>i</code> 个查询的结果。</p>\n\n<p><strong>注意：</strong>节点之间的边是无权重（unweighted）的。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入:</strong> <span class=\"example-io\">n = 5, nums = [1,8,3,4,2], maxDiff = 3, queries = [[0,3],[2,4]]</span></p>\n\n<p><strong>输出:</strong> <span class=\"example-io\">[1,1]</span></p>\n\n<p><strong>解释:</strong></p>\n\n<p>生成的图如下：</p>\n\n<p><img alt=\"\" src=\"https://pic.leetcode.cn/1745660620-PauXMH-4149example1drawio.png\" style=\"width: 281px; height: 161px;\" /></p>\n\n<table>\n\t<tbody>\n\t\t<tr>\n\t\t\t<th>查询</th>\n\t\t\t<th>最短路径</th>\n\t\t\t<th>最短距离</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[0, 3]</td>\n\t\t\t<td>0 → 3</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[2, 4]</td>\n\t\t\t<td>2 → 4</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t</tbody>\n</table>\n\n<p>因此，输出为 <code>[1, 1]</code>。</p>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入:</strong> <span class=\"example-io\">n = 5, nums = [5,3,1,9,10], maxDiff = 2, queries = [[0,1],[0,2],[2,3],[4,3]]</span></p>\n\n<p><strong>输出:</strong> <span class=\"example-io\">[1,2,-1,1]</span></p>\n\n<p><strong>解释:</strong></p>\n\n<p>生成的图如下：</p>\n\n<p><img alt=\"\" src=\"https://pic.leetcode.cn/1745660627-mSVsDs-4149example2drawio.png\" style=\"width: 281px; height: 121px;\" /></p>\n\n<table>\n\t<tbody>\n\t\t<tr>\n\t\t\t<th>查询</th>\n\t\t\t<th>最短路径</th>\n\t\t\t<th>最短距离</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[0, 1]</td>\n\t\t\t<td>0 → 1</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[0, 2]</td>\n\t\t\t<td>0 → 1 → 2</td>\n\t\t\t<td>2</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[2, 3]</td>\n\t\t\t<td>无</td>\n\t\t\t<td>-1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>[4, 3]</td>\n\t\t\t<td>3 → 4</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t</tbody>\n</table>\n\n<p>因此，输出为 <code>[1, 2, -1, 1]</code>。</p>\n</div>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入:</strong> <span class=\"example-io\">n = 3, nums = [3,6,1], maxDiff = 1, queries = [[0,0],[0,1],[1,2]]</span></p>\n\n<p><strong>输出:</strong> <span class=\"example-io\">[0,-1,-1]</span></p>\n\n<p><strong>解释:</strong></p>\n\n<p>由于以下原因，任意两个节点之间都不存在边：</p>\n\n<ul>\n\t<li>节点 0 和节点 1：<code>|nums[0] - nums[1]| = |3 - 6| = 3 &gt; 1</code></li>\n\t<li>节点 0 和节点 2：<code>|nums[0] - nums[2]| = |3 - 1| = 2 &gt; 1</code></li>\n\t<li>节点 1 和节点 2：<code>|nums[1] - nums[2]| = |6 - 1| = 5 &gt; 1</code></li>\n</ul>\n\n<p>因此，不存在任何可以到达其他节点的节点，输出为 <code>[0, -1, -1]</code>。</p>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= maxDiff &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>queries[i] == [u<sub>i</sub>, v<sub>i</sub>]</code></li>\n\t<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt; n</code></li>\n</ul>\n",
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