Connecting Cities With Minimum Cost in Python

Suppose there are N cities numbered from 1 to N. We have connections, where each connection [i] is [city1, city2, cost] this represents the cost to connect city1 and city2 together. We have to find the minimum cost so that for every pair of cities, there exists a path of connections (possibly of length 1) that connects those two cities together. The cost is the sum of the connection costs used. If the task is impossible, return -1.

So if the graph is like −

Then the output will be 6, Choosing any two cities will connect all cities, so we choose the minimum 2, [1, 5]

To solve this, we will follow these steps −

• Define method called find(), this will take x

• if parent[x] is -1, then return x

• parent[x] := find(parent[x])

• return parent[x]

• Make another method called union(), this will take x and y

• parent_x := find(x), parent_y := find(y)

• if parent_x = parent_y, then return, otherwise parent[parent_y] := parent_x

• From the main method, it will take n and conn

• parent := an array of size n and fill this with – 1, set disjoint := n – 1, cost := 0

• c := sorted list of conn, sort this based on the cost

• i := 0

• while i < length of c and disjoint is not 0

  • x := c[i, 0] and y := c[i, 1]

  • if find(x) is not same as find(y), then decrease disjoint by 1, increase the cost by c[i, 2], and perform union(x, y)

  • increase i by 1

• return cost when disjoint is 0, otherwise return -1

Example(Python)

Let us see the following implementation to get a better understanding −

class Solution(object):
   def find(self, x):
      if self.parent[x] == -1:
         return x
      self.parent[x] = self.find(self.parent[x])
      return self.parent[x]
   def union(self,x,y):
      parent_x = self.find(x)
      parent_y = self.find(y)
      if parent_x == parent_y:
         return
      self.parent[parent_y] = parent_x
   def minimumCost(self, n, connections):
      self.parent = [ -1 for i in range(n+1)]
      disjoint = n-1
      cost = 0
      c = sorted(connections,key=lambda v:v[2])
      i = 0
      while i<len(c) and disjoint:
         x = c[i][0]
         y = c[i][1]
         if self.find(x)!=self.find(y):
            disjoint-=1
            cost+=c[i][2]
            self.union(x,y)
            i+=1
         return cost if not disjoint else -1
ob = Solution()
print(ob.minimumCost(3, [[1,2,5], [1,3,6], [2,3,1]]))

Input

3
[[1,2,5],[1,3,6],[2,3,1]]

Output

-1