"content":"<p>There is a <strong>simple directed graph</strong> with <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>. The graph would form a <strong>tree</strong> if its edges were bi-directional.</p>\n\n<p>You are given an integer <code>n</code> and a <strong>2D</strong> integer array <code>edges</code>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> represents a <strong>directed edge</strong> going from node <code>u<sub>i</sub></code> to node <code>v<sub>i</sub></code>.</p>\n\n<p>An <strong>edge reversal</strong> changes the direction of an edge, i.e., a directed edge going from node <code>u<sub>i</sub></code> to node <code>v<sub>i</sub></code> becomes a directed edge going from node <code>v<sub>i</sub></code> to node <code>u<sub>i</sub></code>.</p>\n\n<p>For every node <code>i</code> in the range <code>[0, n - 1]</code>, your task is to <strong>independently</strong> calculate the <strong>minimum</strong> number of <strong>edge reversals</strong> required so it is possible to reach any other node starting from node <code>i</code> through a <strong>sequence</strong> of <strong>directed edges</strong>.</p>\n\n<p>Return <em>an integer array </em><code>answer</code><em>, where </em><code>answer[i]</code><em> is the</em><em> </em> <em><strong>minimum</strong> number of <strong>edge reversals</strong> required so it is possible to reach any other node starting from node </em><code>i</code><em> through a <strong>sequence</strong> of <strong>directed edges</strong>.</em></p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<p><img height=\"246\" src=\"https://assets.leetcode.com/uploads/2023/08/26/image-20230826221104-3.png\" width=\"312\" /></p>\n\n<pre>\n<strong>Input:</strong> n = 4, edges = [[2,0],[2,1],[1,3]]\n<strong>Output:</strong> [1,1,0,2]\n<strong>Explanation:</strong> The image above shows the graph formed by the edges.\nFor node 0: after reversing the edge [2,0], it is possible to reach any other node starting from node 0.\nSo, answer[0] = 1.\nFor node 1: after reversing the edge [2,1], it is possible to reach any other node starting from node 1.\nSo, answer[1] = 1.\nFor node 2: it is already possible to reach any other node starting from node 2.\nSo, answer[2] = 0.\nFor node 3: after reversing the edges [1,3] and [2,1], it is possible to reach any other node starting from node 3.\nSo, answer[3] = 2.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<p><img height=\"217\" src=\"https://assets.leetcode.com/uploads/2023/08/26/image-20230826225541-2.png\" width=\"322\" /></p>\n\n<pre>\n<strong>Input:</strong> n = 3, edges = [[1,2],[2,0]]\n<strong>Output:</strong> [2,0,1]\n<strong>Explanation:</strong> The image above shows the graph formed by the edges.\nFor node 0: after reversing the edges [2,0] and [1,2], it is possible to reach any other node starting from node 0.\nSo, answer[0] = 2.\nFor node 1: it is already possible to reach any other node starting from node 1.\nSo, answer[1] = 0.\nFor node 2: after reversing the edge [1, 2], it is possible to reach any other node starting from node 2.\nSo, answer[2] = 1.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>edges.length == n - 1</code></li>\n\t<li><code>edges[i].length == 2</code></li>\n\t<li><code>0 <= u<sub>i</sub> == edges[i][0] < n</code></li>\n\t<li><code>0 <= v<sub>i</sub> == edges[i][1] < n</code></li>\n\t<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>\n\t<li>The input is generated such that if the edges were bi-directional, the graph would be a tree.</li>\n</ul>\n",
"similarQuestions":"[{\"title\": \"Reorder Routes to Make All Paths Lead to the City Zero\", \"titleSlug\": \"reorder-routes-to-make-all-paths-lead-to-the-city-zero\", \"difficulty\": \"Medium\", \"translatedTitle\": null}]",
"code":"/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* minEdgeReversals(int n, int** edges, int edgesSize, int* edgesColSize, int* returnSize){\n\n}",
"__typename":"CodeSnippetNode"
},
{
"lang":"C#",
"langSlug":"csharp",
"code":"public class Solution {\n public int[] MinEdgeReversals(int n, int[][] edges) {\n \n }\n}",
"Using node <code>0</code> as the root, let <code>dp[x]</code> be the minimum number of edge reversals so node <code>x</code> can reach every node in its subtree.",
"Using a DFS traversing the edges bidirectionally, we can compute <code>dp</code>.<br />\r\n<code>dp[x] = dp[y] +</code> (<code>1</code> if the edge between <code>x</code> and <code>y</code> is going from <code>y</code> to <code>x</code>; <code>0</code> otherwise), where <code>x</code> is the parent of <code>y</code>.",
"Let <code>answer[x]</code> be the minimum number of edge reversals so it is possible to reach any other node starting from node <code>x</code>.",
"Using another DFS starting from node <code>0</code> and traversing the edges bidirectionally, we can compute <code>answer</code>.<br />\r\n<code>answer[0] = dp[0]</code><br />\r\n<code>answer[y] = answer[x] +</code> (<code>1</code> if the edge between <code>x</code> and <code>y</code> is going from <code>x</code> to <code>y</code>; <code>-1</code> otherwise), where <code>x</code> is the parent of <code>y</code>."
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