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Author: jimit105

2438-find-closest-node-to-given-two-nodes/README.md

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+<p>You are given a <strong>directed</strong> graph of <code>n</code> nodes numbered from <code>0</code> to <code>n - 1</code>, where each node has <strong>at most one</strong> outgoing edge.</p>
+
+<p>The graph is represented with a given <strong>0-indexed</strong> array <code>edges</code> of size <code>n</code>, indicating that there is a directed edge from node <code>i</code> to node <code>edges[i]</code>. If there is no outgoing edge from <code>i</code>, then <code>edges[i] == -1</code>.</p>
+
+<p>You are also given two integers <code>node1</code> and <code>node2</code>.</p>
+
+<p>Return <em>the <strong>index</strong> of the node that can be reached from both </em><code>node1</code><em> and </em><code>node2</code><em>, such that the <strong>maximum</strong> between the distance from </em><code>node1</code><em> to that node, and from </em><code>node2</code><em> to that node is <strong>minimized</strong></em>. If there are multiple answers, return the node with the <strong>smallest</strong> index, and if no possible answer exists, return <code>-1</code>.</p>
+
+<p>Note that <code>edges</code> may contain cycles.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2022/06/07/graph4drawio-2.png" style="width: 321px; height: 161px;" />
+<pre>
+<strong>Input:</strong> edges = [2,2,3,-1], node1 = 0, node2 = 1
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> The distance from node 0 to node 2 is 1, and the distance from node 1 to node 2 is 1.
+The maximum of those two distances is 1. It can be proven that we cannot get a node with a smaller maximum distance than 1, so we return node 2.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2022/06/07/graph4drawio-4.png" style="width: 195px; height: 161px;" />
+<pre>
+<strong>Input:</strong> edges = [1,2,-1], node1 = 0, node2 = 2
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> The distance from node 0 to node 2 is 2, and the distance from node 2 to itself is 0.
+The maximum of those two distances is 2. It can be proven that we cannot get a node with a smaller maximum distance than 2, so we return node 2.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == edges.length</code></li>
+	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>-1 &lt;= edges[i] &lt; n</code></li>
+	<li><code>edges[i] != i</code></li>
+	<li><code>0 &lt;= node1, node2 &lt; n</code></li>
+</ul>

2438-find-closest-node-to-given-two-nodes/solution.py

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+# Approach 2: Depth First Search
+
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def dfs(self, node, edges, dist, visit):
+        visit[node] = True
+        neighbor = edges[node]
+        if neighbor != -1 and not visit[neighbor]:
+            dist[neighbor] = dist[node] + 1
+            self.dfs(neighbor, edges, dist, visit)
+
+
+    def closestMeetingNode(self, edges: List[int], node1: int, node2: int) -> int:
+        n = len(edges)
+        dist1 = [float('inf')] * n
+        dist2 = [float('inf')] * n
+        dist1[node1] = 0
+        dist2[node2] = 0
+        
+        visit1 = [False] * n
+        visit2 = [False] * n
+
+        self.dfs(node1, edges, dist1, visit1)
+        self.dfs(node2, edges, dist2, visit2)
+
+        minDistNode = -1
+        minDistTillNow = float('inf')
+
+        for currNode in range(n):
+            if minDistTillNow > max(dist1[currNode], dist2[currNode]):
+                minDistNode = currNode
+                minDistTillNow = max(dist1[currNode], dist2[currNode])
+
+        return minDistNode