feat: add lc 684

apps/docs/content/union-find/684-redundant-connection.mdx

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+---
+title: '684. Redundant Connection'
+description: In this problem, a tree is an undirected graph that is connected and has no cycles
+sidebar:
+  label: 'Redundant Connection'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Depth-First Search</Badge>
+
+### Example 1:
+- Input: `edges = [[1,2],[1,3],[2,3]]`
+- Output: `[2,3]`
+
+### Example 2:
+- Input: `edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]`
+- Output: `[1,4]`
+
+### Constraints:
+
+- `n == edges.length`
+- `3 <= n <= 1000`
+- `edges[i].length == 2`
+- `1 <= ai < bi <= edges.length`
+- `ai != bi`
+- There are no repeated edges.
+- The given graph is connected.
+
+## Solution
+
+```py
+class Solution:
+    def findRedundantConnection(self, edges: list[list[int]]) -> list[int]:
+        N = len(edges)
+        par = [i for i in range(N + 1)]
+        rank = [1] * (N + 1)
+
+        def find(n):
+            if n != par[n]:
+                par[n] = find(par[n])
+            return par[n]
+
+        def union(n1, n2):
+            p1, p2 = find(n1), find(n2)
+            if p1 == p2:
+                return False
+
+            if rank[p1] > rank[p2]:
+                par[p2] = p1
+                rank[p1] += rank[p2]
+            else:
+                par[p1] = p2
+                rank[p2] += rank[p1]
+
+            return True
+
+        for n1, n2 in edges:
+            if not union(n1, n2):
+                return [n1, n2]
+```

apps/docs/content/union-find/index.mdx

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+---
+title: Union Find
+description: 'Union-Find (Disjoint Set Union) is a data structure that maintains a collection of disjoint sets and supports efficiently finding which set an element belongs to and merging two sets.'
+hidden: true
+---
+
+
+
+

apps/docs/content/union-find/meta.ts

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+import { defineMeta } from "blume";
+
+export default defineMeta({ title: "Union-Find", order: 19 });

work/1/Medium/Depth-First Search/684.redundant-connection.py

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+"""
+684. Redundant Connection
+Difficulty: Medium
+https://leetcode.com/problems/redundant-connection/
+
+──────────────────────────────────────────────────
+
+In this problem, a tree is an undirected graph that is connected and
+has no cycles.
+
+You are given a graph that started as a tree with n nodes labeled
+from 1 to n, with one additional edge added. The added edge has two
+different vertices chosen from 1 to n, and was not an edge that
+already existed. The graph is represented as an array edges of length
+n where edges[i] = [ai, bi] indicates that there is an edge between
+nodes ai and bi in the graph.
+
+Return an edge that can be removed so that the resulting graph is a
+tree of n nodes. If there are multiple answers, return the answer that
+occurs last in the input.
+
+
+
+Example 1:
+
+Input: edges = [[1,2],[1,3],[2,3]]
+Output: [2,3]
+
+Example 2:
+
+Input: edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]
+Output: [1,4]
+
+
+
+Constraints:
+
+        • n == edges.length
+
+        • 3 <= n <= 1000
+
+        • edges[i].length == 2
+
+        • 1 <= ai < bi <= edges.length
+
+        • ai != bi
+
+        • There are no repeated edges.
+
+        • The given graph is connected.
+"""
+
+
+class Solution:
+    def findRedundantConnection(self, edges: list[list[int]]) -> list[int]:
+        N = len(edges)
+        par = [i for i in range(N + 1)]
+        rank = [1] * (N + 1)
+
+        def find(n):
+            if n != par[n]:
+                par[n] = find(par[n])
+            return par[n]
+
+        def union(n1, n2):
+            p1, p2 = find(n1), find(n2)
+            if p1 == p2:
+                return False
+
+            if rank[p1] > rank[p2]:
+                par[p2] = p1
+                rank[p1] += rank[p2]
+            else:
+                par[p1] = p2
+                rank[p2] += rank[p1]
+
+            return True
+
+        for n1, n2 in edges:
+            if not union(n1, n2):
+                return [n1, n2]