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 −

Example

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.append([0,i+1,well[i]])
         counter+=1
      pipes = sorted(pipes,key=lambda v:v[2])
      cost = 0
      for i in pipes:
         #print(i)
         source = i[0]
         destination = i[1]
         temp = i[2]
         if not self.union(source,destination):
            cost+=temp
      return cost

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

Input

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

Optimize Water Distribution in a Village in Python

Example

Input

Output