Sync LeetCode submission Runtime - 11 ms (17.39%), Memory - 18.2 MB (51.11%)

Author: jimit105

0684-redundant-connection/README.md

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+<p>In this problem, a tree is an <strong>undirected graph</strong> that is connected and has no cycles.</p>
+
+<p>You are given a graph that started as a tree with <code>n</code> nodes labeled from <code>1</code> to <code>n</code>, with one additional edge added. The added edge has two <strong>different</strong> vertices chosen from <code>1</code> to <code>n</code>, and was not an edge that already existed. The graph is represented as an array <code>edges</code> of length <code>n</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the graph.</p>
+
+<p>Return <em>an edge that can be removed so that the resulting graph is a tree of </em><code>n</code><em> nodes</em>. If there are multiple answers, return the answer that occurs last in the input.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/05/02/reduntant1-1-graph.jpg" style="width: 222px; height: 222px;" />
+<pre>
+<strong>Input:</strong> edges = [[1,2],[1,3],[2,3]]
+<strong>Output:</strong> [2,3]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/05/02/reduntant1-2-graph.jpg" style="width: 382px; height: 222px;" />
+<pre>
+<strong>Input:</strong> edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]
+<strong>Output:</strong> [1,4]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == edges.length</code></li>
+	<li><code>3 &lt;= n &lt;= 1000</code></li>
+	<li><code>edges[i].length == 2</code></li>
+	<li><code>1 &lt;= a<sub>i</sub> &lt; b<sub>i</sub> &lt;= edges.length</code></li>
+	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
+	<li>There are no repeated edges.</li>
+	<li>The given graph is connected.</li>
+</ul>

0684-redundant-connection/solution.py

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+# Approach 1: Depth-First Search - Brute Force
+
+# Time: O(n ^ 2)
+# Space: O(n)
+
+class Solution:
+    def _is_connected(self, src, target, visited, adj_list):
+        visited[src] = True
+
+        if src == target:
+            return True
+        
+        is_found = False
+        for adj in adj_list[src]:
+            if not visited[adj]:
+                is_found = is_found or self._is_connected(adj, target, visited, adj_list)
+
+        return is_found
+
+
+    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
+        n = len(edges)
+
+        adj_list = [[] for _ in range(n)]
+
+        for edge in edges:
+            visited = [False] * n
+
+            if self._is_connected(edge[0] - 1, edge[1] - 1, visited, adj_list):
+                return edge
+
+            adj_list[edge[0] - 1].append(edge[1] - 1)
+            adj_list[edge[1] - 1].append(edge[0] - 1)
+
+        return []