solves Count Unreachable Pairs of Nodes in an Undirected Graph (#2316) in python (#12)

Author: sumanth-botlagunta

python/count_unreachable_pair_of_node_in_an_undirectable_graph.py

@@ -0,0 +1,49 @@
+# https://leetcode.com/problems/count-unreachable-pairs-of-nodes-in-an-undirected-graph/description/
+# T: O(N + E) where N is the number of nodes and E is the number of edges
+# S: O(N + E) where N is the number of nodes and E is the number of edges
+
+from collections import defaultdict
+class Solution:
+    def __init__(self):
+        self.graph = defaultdict(list)
+        self.visited = set()
+        self.n = 0
+
+    def add_edge(self, u, v):
+        self.graph[u].append(v)
+        self.graph[v].append(u)
+
+    def dfs(self, v, component):
+        self.visited.add(v)
+        component.append(v)
+        for neighbor in self.graph[v]:
+            if neighbor not in self.visited:
+                self.dfs(neighbor, component)
+
+    def find_components(self):
+        components = []
+        for v in range(self.n):
+            if v not in self.visited:
+                component = []
+                self.dfs(v, component)
+                components.append(component)
+
+        return components
+
+    def countPairs(self, n, edges):
+        for u, v in edges:
+            self.add_edge(u, v)
+        self.n = n
+        component_lengths = []
+        components = self.find_components()
+        for component in components:
+            component_lengths.append(len(component))
+        print(components, component_lengths)
+
+        k = sum(component_lengths)
+        sol = 0
+        for l in component_lengths:
+            sol+=((k-l)*l)
+        sol = sol//2
+
+        return sol