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Commit 9bb99f5

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Added medium problem.
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"""
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# NUMBER OF PROVINCES
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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.
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A province is a group of directly or indirectly connected cities and no other cities outside of the group.
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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.
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Return the total number of provinces.
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Example 1:
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Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]]
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Output: 2
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Example 2:
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Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]]
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Output: 3
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Constraints:
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1 <= n <= 200
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n == isConnected.length
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n == isConnected[i].length
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isConnected[i][j] is 1 or 0.
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isConnected[i][i] == 1
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isConnected[i][j] == isConnected[j][i]
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"""
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class Solution:
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def findCircleNum(self, isConnected):
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graph = {}
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n = len(isConnected)
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for i in range(n):
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graph[i] = []
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for j in range(n):
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if isConnected[i][j] == 1:
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graph[i].append(j)
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print(graph)
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provinces = 0
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visited = set()
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for i in range(n):
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if i not in visited:
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provinces += 1
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self.runDfs(i, graph, visited)
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return provinces
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def runDfs(self, current, graph, visited):
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if current in visited:
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return
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visited.add(current)
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for neighbor in graph[current]:
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self.runDfs(neighbor, graph, visited)
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return

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