chore: clean up content phase organization

apps/docs/content/(04-relational)/union-find/323-number-of-connected-components-in-an-undirected-graph.mdx

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+---
+title: '323. Number of Connected Components in an Undirected Graph'
+description: You have a graph of n nodes. You are given an integer n and an array edges where edges[i] = [ai, bi] indicates that there is an edge between ai and bi in the graph
+sidebar:
+  label: 'Number of Connected Components in an Undirected Graph'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Depth-First Search</Badge>
+
+### Example 1:
+- Input: `n = 5, edges = [[0,1],[1,2],[3,4]]`
+- Output: `2`
+
+### Example 2:
+- Input: `n = 5, edges = [[0,1],[1,2],[2,3],[3,4]]`
+- Output: `1`
+
+### Constraints:
+
+- `1 <= n <= 2000`
+- `1 <= edges.length <= 5000`
+- `edges[i]` = [ai, bi]
+- `ai != bi`
+- There are no repeated edges.
+
+## Solution
+
+```py
+class Solution:
+    def countComponents(self, n: int, edges: list[list[int]]) -> int:
+        par = [i for i in range(n + 1)]
+        rank = [1] * (n + 1)
+
+        def find(n):
+            res = n
+            while res != par[res]:
+                par[res] = par[par[res]]
+                res = par[res]
+            return res
+
+        def union(n1, n2):
+            p1, p2 = find(n1), find(n2)
+            if p1 == p2:
+                return 0
+            if rank[p1] > rank[p2]:
+                par[p2] = p1
+                rank[p1] += rank[p2]
+            else:
+                par[p1] = p2
+                rank[p2] += rank[p1]
+            return 1
+
+        res = n
+        for n1, n2 in edges:
+            res -= union(n1, n2)
+        return res
+```

apps/docs/content/(04-relational)/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/(04-relational)/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/(04-relational)/union-find/meta.ts

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