Feat: 323. Number of Connected Components in an Undirected Graph

Author: HC-kang

number-of-connected-components-in-an-undirected-graph/HC-kang.ts

@@ -0,0 +1,71 @@
+/**
+ * https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph
+ * T.C. O(V + E)
+ * S.C. O(V + E)
+ */
+function countComponents(n: number, edges: number[][]): number {
+  const graph = new Map<number, number[]>();
+
+  for (let i = 0; i < n; i++) {
+    graph.set(i, []);
+  }
+
+  for (const [u, v] of edges) {
+    graph.get(u)!.push(v);
+    graph.get(v)!.push(u);
+  }
+
+  function dfs(node: number) {
+    visited.add(node);
+
+    for (const neighbor of graph.get(node)!) {
+      if (!visited.has(neighbor)) {
+        dfs(neighbor);
+      }
+    }
+  }
+
+  const visited = new Set<number>();
+
+  let components = 0;
+
+  for (let i = 0; i < n; i++) {
+    if (!visited.has(i)) {
+      components++;
+      dfs(i);
+    }
+  }
+
+  return components;
+}
+
+/**
+ * Using Union Find
+ * T.C. O(V + E)
+ *
+ */
+function countComponents(n: number, edges: number[][]): number {
+  const parent = Array.from({ length: n }, (_, i) => i);
+
+  // find and compress path
+  function find(x: number): number {
+    if (parent[x] !== x) {
+      parent[x] = find(parent[x]);
+    }
+    return parent[x];
+  }
+
+  // union two sets and check if they have the same root
+  function union(x: number, y: number): boolean {
+    const rootX = find(x);
+    const rootY = find(y);
+    parent[rootX] = rootY;
+    return true;
+  }
+
+  for (const [x, y] of edges) {
+    union(x, y);
+  }
+
+  return new Set(parent.map(find)).size;
+}