feat(work): add solution for LeetCode 323 number of connected components problem

work/1/Medium/Depth-First Search/323.number-of-connected-components-in-an-undirected-graph.py

@@ -0,0 +1,69 @@
+"""
+323. Number of Connected Components in an Undirected Graph
+Difficulty: Medium
+https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/
+
+──────────────────────────────────────────────────
+
+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.
+
+Return the number of connected components in the graph.
+
+
+
+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.
+"""
+
+
+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