Program to count minimum number of animals which have no predator in Python

Suppose we have a list of numbers called nums where nums[i] shows the predator of the ith animal and if there is no predator, it will hold −1. We have to find the smallest number of groups of animals such that no animal is in the same group with its direct or indirect predator.

So, if the input is like nums = [1, 2, −1, 4, 5, −1], then the output will be 3, as we can have the groups like: [0, 3], [1, 4], [2, 5].

To solve this, we will follow these steps −

• if A is empty, then

  • return 0

• adj := a blank map

• vis := a new set

• roots := a new list

• for each index i and value a in A, do

  • if a is same as −1, then

    • insert i at the end of roots

  • insert a at the end of adj[i]

  • insert i at the end of adj[a]

• best := −infinity

• for each root in roots, do

  • stk := a stack and insert [root, 1] into it

  • while stk is not empty, do

    • (node, d) := popped element of stk

    • if node is in vis or node is same as −1, then

      • come out from the loop

    • best := maximum of best and d

    • insert node into vis

    • for each u in adj[node], do

      • push (u, d + 1) into stk

• return best

Let us see the following implementation to get better understanding −

Example

 Live Demo

from collections import defaultdict class Solution:    def solve(self, A):       if not A:          return 0       adj = defaultdict(list)       vis = set()       roots = []       for i, a in enumerate(A):          if a == -1:             roots.append(i)          adj[i].append(a)          adj[a].append(i)       best = −float("inf")       for root in roots:       stk = [(root, 1)]       while stk:          node, d = stk.pop()          if node in vis or node == −1:             continue          best = max(best, d)          vis.add(node)          for u in adj[node]:             stk.append((u, d + 1))    return best ob = Solution() nums = [1, 2, −1, 4, 5, −1] print(ob.solve(nums))

Input

[1, 2, −1, 4, 5, −1]

Output

3