Program to find shortest cycle length holding target in python

Suppose we have adjacency list of a directed graph, where each list at index i is represented connected nodes from node i. We also have a target value. We have to find the length of a shortest cycle that contains the target. If there is no solution return -1.

So, if the input is like

target = 3., then the output will be 3, as the cycle is nodes 1 -> 2 -> 3 -> 1. Note there is another cycle 0 -> 1 -> 2 -> 3 -> 0, but this is not shortest.

To solve this, we will follow these steps -

• visited := a new set
• l := a list with element target.
• length := 0
• while l is non-empty, do
  • length := length + 1
  • nl := a new list
  • for each u in l, do
    • for each v in graph[u], do
      • if v is same as target, then
        • return length
      • if v is visited, then
        • go for the next iteration
      • mark v as visited
      • insert v at the end of nl
  • l := nl
• return -1

Let us see the following implementation to get better understanding -

Example 

Live Demo

class Solution:
   def solve(self, graph, target):
      visited = set()
      l = [target]
      length = 0

      while l:
         length += 1
         nl = []
         for u in l:
            for v in graph[u]:
               if v == target:
                  return length
               if v in visited:
                  continue
               visited.add(v)
               nl.append(v)
         l = nl

      return -1

ob = Solution()
graph = [[1, 4],[2],[3],[0, 1],[]]
target = 3
print(ob.solve(graph, target))

Input

[[1, 4],[2],[3],[0, 1],[]]

Output

3

Arnab Chakraborty

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Updated on: 02-Dec-2020

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