Sync LeetCode submission Runtime - 1 ms (58.70%), Memory - 18.9 MB (69.09%)

Author: jimit105

0261-graph-valid-tree/README.md

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+<p>You have a graph of <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>. You are given an integer n and a list of <code>edges</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an undirected edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the graph.</p>
+
+<p>Return <code>true</code> <em>if the edges of the given graph make up a valid tree, and</em> <code>false</code> <em>otherwise</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/03/12/tree1-graph.jpg" style="width: 222px; height: 302px;" />
+<pre>
+<strong>Input:</strong> n = 5, edges = [[0,1],[0,2],[0,3],[1,4]]
+<strong>Output:</strong> true
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/03/12/tree2-graph.jpg" style="width: 382px; height: 222px;" />
+<pre>
+<strong>Input:</strong> n = 5, edges = [[0,1],[1,2],[2,3],[1,3],[1,4]]
+<strong>Output:</strong> false
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 2000</code></li>
+	<li><code>0 &lt;= edges.length &lt;= 5000</code></li>
+	<li><code>edges[i].length == 2</code></li>
+	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
+	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
+	<li>There are no self-loops or repeated edges.</li>
+</ul>

0261-graph-valid-tree/solution.py

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+# Approach 2: Advanced Graph Theory + Iterative Depth-First Search
+
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def validTree(self, n: int, edges: List[List[int]]) -> bool:
+        if len(edges) != (n - 1):
+            return False
+
+        adj_list = [[] for _ in range(n)]
+        for a, b in edges:
+            adj_list[a].append(b)
+            adj_list[b].append(a)
+
+        seen = {0}
+        stack = [0]
+
+        while stack:
+            node = stack.pop()
+            for neighbor in adj_list[node]:
+                if neighbor in seen:
+                    continue
+                seen.add(neighbor)
+                stack.append(neighbor)
+
+        return len(seen) == n
+