solve(w12): 323. Number of Connected Components in an Undirected Graph

Author: seungriyou

number-of-connected-components-in-an-undirected-graph/seungriyou.py

@@ -0,0 +1,108 @@
+# https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/
+
+from typing import List
+
+class Solution:
+    def countComponents_bfs(self, n: int, edges: List[List[int]]) -> int:
+        """
+        [Complexity]
+            - TC: O(v + e) (모든 edge & node 한 번씩 탐색)
+            - SC: O(v + e) (graph)
+
+        [Approach]
+            BFS로 visited를 기록해가며 connected component를 센다.
+        """
+        from collections import deque
+
+        graph = [[] for _ in range(n)]
+        for a, b in edges:
+            graph[a].append(b)
+            graph[b].append(a)
+
+        visited, res = set(), 0
+
+        def bfs(start):
+            q = deque([start])
+            visited.add(start)
+
+            while q:
+                pos = q.popleft()
+                for npos in graph[pos]:
+                    if npos not in visited:
+                        q.append(npos)
+                        visited.add(npos)
+
+            return
+
+        for i in range(n):
+            if i not in visited:
+                bfs(i)
+                res += 1
+
+        return res
+
+    def countComponents_dfs(self, n: int, edges: List[List[int]]) -> int:
+        """
+        [Complexity]
+            - TC: O(v + e) (모든 edge & node 한 번씩 탐색)
+            - SC: O(v + e) (graph)
+
+        [Approach]
+            DFS로 visited를 기록해가며 connected component를 센다.
+        """
+        graph = [[] for _ in range(n)]
+        for a, b in edges:
+            graph[a].append(b)
+            graph[b].append(a)
+
+        visited, res = set(), 0
+
+        def dfs(pos):
+            # 이전에 visited 포함 여부 확인하므로 base condition 생략 가능
+
+            visited.add(pos)
+
+            # recur
+            for npos in graph[pos]:
+                if npos not in visited:
+                    dfs(npos)
+
+            return
+
+        for i in range(n):
+            if i not in visited:
+                dfs(i)
+                res += 1
+
+        return res
+
+    def countComponents(self, n: int, edges: List[List[int]]) -> int:
+        """
+        [Complexity]
+            - TC: O(v + e * α(v)) (모든 edge & node 한 번씩 탐색)
+            - SC: O(v) (parent, set(...))
+
+        [Approach]
+            edges를 iterate 하며 union-find 수행 후, parent의 종류의 개수를 세면 된다.
+            parent의 종류의 개수를 셀 때는 다시 find_parent(x)로 찾아야 한다!
+        """
+
+        def find_parent(x):
+            if x != parent[x]:
+                parent[x] = find_parent(parent[x])
+            return parent[x]
+
+        def union_parent(x, y):
+            px, py = find_parent(x), find_parent(y)
+
+            if px < py:
+                parent[py] = px
+            else:
+                parent[px] = py
+
+        parent = [i for i in range(n)]
+
+        for x, y in edges:
+            union_parent(x, y)
+
+        return len(set(find_parent(i) for i in range(n)))