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58 lines
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58 lines
3.6 KiB
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<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>
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<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>
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<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>
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<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>
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<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>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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<p><img height="246" src="https://assets.leetcode.com/uploads/2023/08/26/image-20230826221104-3.png" width="312" /></p>
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<pre>
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<strong>Input:</strong> n = 4, edges = [[2,0],[2,1],[1,3]]
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<strong>Output:</strong> [1,1,0,2]
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<strong>Explanation:</strong> The image above shows the graph formed by the edges.
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For node 0: after reversing the edge [2,0], it is possible to reach any other node starting from node 0.
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So, answer[0] = 1.
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For node 1: after reversing the edge [2,1], it is possible to reach any other node starting from node 1.
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So, answer[1] = 1.
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For node 2: it is already possible to reach any other node starting from node 2.
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So, answer[2] = 0.
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For node 3: after reversing the edges [1,3] and [2,1], it is possible to reach any other node starting from node 3.
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So, answer[3] = 2.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<p><img height="217" src="https://assets.leetcode.com/uploads/2023/08/26/image-20230826225541-2.png" width="322" /></p>
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<pre>
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<strong>Input:</strong> n = 3, edges = [[1,2],[2,0]]
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<strong>Output:</strong> [2,0,1]
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<strong>Explanation:</strong> The image above shows the graph formed by the edges.
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For node 0: after reversing the edges [2,0] and [1,2], it is possible to reach any other node starting from node 0.
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So, answer[0] = 2.
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For node 1: it is already possible to reach any other node starting from node 1.
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So, answer[1] = 0.
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For node 2: after reversing the edge [1, 2], it is possible to reach any other node starting from node 2.
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So, answer[2] = 1.
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>2 <= n <= 10<sup>5</sup></code></li>
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<li><code>edges.length == n - 1</code></li>
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<li><code>edges[i].length == 2</code></li>
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<li><code>0 <= u<sub>i</sub> == edges[i][0] < n</code></li>
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<li><code>0 <= v<sub>i</sub> == edges[i][1] < n</code></li>
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<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
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<li>The input is generated such that if the edges were bi-directional, the graph would be a tree.</li>
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</ul>
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