added Number of Operations to Make Network Connected (medium)

Author: minhyeong-joe

Medium/NumToMakeNetworkConnected/README.md

@@ -0,0 +1,34 @@
+
+# Number of Operations to Make Network Connected
+[Leetcode Link](https://leetcode.com/problems/number-of-operations-to-make-network-connected/)
+
+## Problem:
+
+There are `n` computers numbered from `0` to `n-1` connected by ethernet cables connections forming a network where `connections[i] = [a, b]` represents a connection between computers `a` and `b`. Any computer can reach any other computer directly or indirectly through the network.
+
+Given an initial computer network `connections`. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected. Return the *minimum number* of times you need to do this in order to make all the computers connected. If it's not possible, return -1. 
+
+## Example:
+
+![example](assets/example1.png)
+```
+Input: n = 4, connections = [[0,1],[0,2],[1,2]]
+Output: 1
+Explanation: Remove cable between computer 1 and 2 and place between computers 1 and 3.
+```
+![example](assets/example2.png)
+```
+Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2],[1,3]]
+Output: 2
+```
+```
+Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2]]
+Output: -1
+Explanation: There are not enough cables.
+```
+```
+Input: n = 5, connections = [[0,1],[0,2],[3,4],[2,3]]
+Output: 0
+```
+
+## Note:

Medium/NumToMakeNetworkConnected/assets/example1.png

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+from typing import List
+from collections import defaultdict
+
+
+class Solution:
+ # the main idea is that if number of connections is less than number of computers - 1, then it is not possible to form connections
+ # else, find how many disjoint graphs there are, and the number of connections that needs to be modified is simply num of disjoint - 1
+ def makeConnected(self, n: int, connections: List[List[int]]) -> int:
+ if len(connections) < n-1:
+ return -1
+ undirectedGraph = defaultdict(list)
+ for a, b in connections:
+ undirectedGraph[a].append(b)
+ undirectedGraph[b].append(a)
+ # print(undirectedGraph)
+ visited = set()
+ queue = list()
+ numDisjoint = 0
+ # visit all computers (even disjoint ones)
+ for i in range(n):
+ # if computer i has been visited by BFS, then do not increment number of disjoint graph
+ if i not in visited:
+ queue.append(i)
+ # the ones in the same queue belong to the same graph group
+ while queue:
+ current = queue.pop(0)
+ visited.add(current)
+ for neighbor in undirectedGraph[current]:
+ if neighbor not in visited and neighbor not in queue:
+ queue.append(neighbor)
+ numDisjoint += 1
+ return numDisjoint - 1
+
+
+# test driver
+sol = Solution()
+n = 9
+connections = [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [4, 5], [4, 6], [5, 6]]
+print("Output:", sol.makeConnected(n, connections))

Medium/NumToMakeNetworkConnected/assets/example2.png

@@ -72,6 +72,7 @@ Languages used: Java and Python
  - [Subarray Sum Equals K](Medium/SubarraySumEqualsK)
  - [Find Minimum in Rotated Sorted Array](Medium/FindMinInRotatedSortedArray)
  - [Maximum Swap](Medium/MaximumSwap)
+ - [Number of Operations to Make Network Connected](Medium/NumToMakeNetworkConnected)
 - Hard
  - [Maximum Score Words Formed by Letters](Hard/MaximumScoreWords)
  - [Reducing Dishes](Hard/ReducingDishes)