Create 2025-04-24-Leetcode-684.md

Author: hyeonukim

_posts/Leetcode/Graphs/2025-04-24-Leetcode-684.md

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+---
+title: Leetcode 684. Redundant Connection
+description: Explanation for Leetcode 684 - Redundant Connection, and its solution in Python.
+date: 2025-04-24
+categories: [Leetcode, Graphs, Medium]
+tags: [Leetcode, Python, Study, Graphs, Medium]
+math: true
+---
+
+## Problem
+[Leetcode 684 - Redundant Connection](https://leetcode.com/problems/redundant-connection/description/)
+
+Example:
+```
+Input: edges = [[1,2],[1,3],[2,3]]
+Output: [2,3]
+
+Input: edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]
+Output: [1,4]
+```
+
+## Approach
+
+For a graph to have a cycle, every node must have 2 or more neighbors, thus we can use tropological sort to get all the neighbor counts(indegree counts) then if indegree == 1, we can disconnect them first. Once we're all done with the process of disconnecting the 1 neighbor edge, we can loop through the edges.
+
+If the indgree count is == 2, and the adjacent indegree count exists, then we can return that edge.
+
+Here is the Python code for the solution:
+```python
+class Solution:
+    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
+        n = len(edges)
+        indegree = [0] * (n+1)
+        adj = [[] for _ in range(n+1)]
+
+        for u, v in edges:
+            adj[u].append(v)
+            adj[v].append(u)
+            indegree[u] += 1
+            indegree[v] += 1
+        
+        q = deque()
+        for i in range(1, n+1):
+            if indegree[i] == 1:
+                q.append(i)
+
+        while q:
+            node = q.popleft()
+
+            indegree[node] -= 1
+            for i in adj[node]:
+                indegree[i] -= 1
+                if indegree[i] == 1:
+                    q.append(i)
+        
+        for u, v in reversed(edges):
+            if indegree[u] == 2 and indegree[v]:
+                return [u, v]
+        
+        return []    
+```
+## Time Complexity and Space Complexity
+
+Time Complexity: $O(V + E)$
+
+Space Complexity: $O(V + E)$