Optimize Water Distribution in a Village in Python

Suppose there are n houses in a village. We have to supply water for all the houses by building wells and laying pipes. For each house i, we can either build a well inside it, the building cost will be wells[i], or pipe in water from another well to it. The costs to lay pipes between houses are given by the array pipes, where each pipes[i] is [house1, house2, cost] represents the cost to connect house1 and house2 together using a pipe. These connections are bidirectional. We have to find the minimum total cost to supply water to all houses.

So, if the input is like n = 3, wells = [1,2,2], pipes = [[1,2,1],[2,3,1]], then the output will be 3

as from the above image shows the costs of connecting houses using pipes. The best strategy will be to build a well in the first house with cost 1 and connect the other houses to it with cost 2 so the total cost is 3.

To solve this, we will follow these steps −

  • Define a function find(). This will take a

  • if parent[a] is same as -1, then

    • return a

  • parent[a] := find(parent[a])

  • return parent[a]

  • Define a function union(). This will take a,b

  • parent_a := find(a)

  • parent_b := find(b)

  • if parent_a is same as parent_b, then

    • return True

  • parent[parent_b] := parent_a

  • return False

  • From the main method do the following −

  • parent := make a list of size n + 1, fill this with -1

  • counter := 0

  • for i in range 0 to size of well, do

    • insert [0, i+1, well[i]] at the end of pipes

    • counter := counter + 1

  • sort the pipes array based on the cost

  • cost := 0

  • for each i in pipes, do

    • source := i[0]

    • destination := i[1]

    • temp := i[2]

    • if union(source,destination) is false, then

      • cost := cost + temp

  • return cost

Let us see the following implementation to get better understanding −


 Live Demo

class Solution(object):
   def find(self, a):
      if self.parent[a] == -1:
         return a
      self.parent[a] = self.find(self.parent[a])
      return self.parent[a]
   def union(self,a,b):
      parent_a = self.find(a)
      parent_b = self.find(b)
      if parent_a == parent_b:
         return True
      self.parent[parent_b] = parent_a
      return False
   def minCostToSupplyWater(self, n, well, pipes):
      self.parent = [-1 for i in range(n+1)]
      counter = 0
      for i in range(len(well)):
      pipes = sorted(pipes,key=lambda v:v[2])
      cost = 0
      for i in pipes:
         source = i[0]
         destination = i[1]
         temp = i[2]
         if not self.union(source,destination):
      return cost

ob = Solution()
print(ob.minCostToSupplyWater(3, [1,2,2], [[1,2,1],[2,3,1]]))


3, [1,2,2], [[1,2,1],[2,3,1]]



Updated on: 11-Jul-2020


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