Added medium problem.

Author: das-jishu

Leetcode/medium/number-of-provinces.py

@@ -0,0 +1,57 @@
+"""
+# NUMBER OF PROVINCES
+
+There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c.
+
+A province is a group of directly or indirectly connected cities and no other cities outside of the group.
+
+You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the ith city and the jth city are directly connected, and isConnected[i][j] = 0 otherwise.
+
+Return the total number of provinces.
+
+Example 1:
+
+Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]]
+Output: 2
+Example 2:
+
+Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]]
+Output: 3
+
+Constraints:
+
+1 <= n <= 200
+n == isConnected.length
+n == isConnected[i].length
+isConnected[i][j] is 1 or 0.
+isConnected[i][i] == 1
+isConnected[i][j] == isConnected[j][i]
+"""
+
+class Solution:
+    def findCircleNum(self, isConnected):
+        graph = {}
+        n = len(isConnected)
+        for i in range(n):
+            graph[i] = []
+            for j in range(n):
+                if isConnected[i][j] == 1:
+                    graph[i].append(j)
+        print(graph)            
+        provinces = 0
+        visited = set()
+        for i in range(n):
+            if i not in visited:
+                provinces += 1
+                self.runDfs(i, graph, visited)
+        
+        return provinces
+        
+    def runDfs(self, current, graph, visited):
+        if current in visited:
+            return
+        
+        visited.add(current)
+        for neighbor in graph[current]:
+            self.runDfs(neighbor, graph, visited)
+        return