Program to separate persons where no enemies can stay in same group in Python

PythonServer Side ProgrammingProgramming

Suppose we have a number n and a 2D matrix called enemies. Here n indicates there is n people labeled from [0, n - 1]. Now each row in enemies contains [a, b] which means that a and b are enemies. We have to check whether it is possible to partition the n people into two groups such that no two people that are enemies are in the same group.

So, if the input is like n = 4, enemies = [[0, 3],[3, 2]], then the output will be True, as we can have these two groups [0, 1, 2] and [3].

To solve this, we will follow these steps −

  • graph := an empty adjacency list

  • for each enemy pair(u, v) in enemies, do

    • insert v at the end of graph[u]

    • insert u at the end of graph[v]

  • color := a new map

  • Define a function dfs() . This will take u, c := 0 initially

  • if u is in color, then

    • return true when color[u] is same as c

  • color[u] := c

  • return true when all of dfs(v, c XOR 1) for each v in graph[u] are true

  • From the main method do the following −

  • return true when all of (dfs(u) for each u in range 0 to n and if u is not in color) are true

Let us see the following implementation to get better understanding −

Example

 Live Demo

class Solution:
   def solve(self, n, enemies):
      from collections import defaultdict
      graph = defaultdict(list)
      for u, v in enemies:
         graph[u].append(v)
         graph[v].append(u)
      color = {}
      def dfs(u, c=0):
         if u in color:
            return color[u] == c
            color[u] = c
            return all(dfs(v, c ^ 1) for v in graph[u])
         return all(dfs(u) for u in range(n) if u not in color)
ob = Solution()
n = 4
enemies = [[0, 3],[3, 2]]
print(ob.solve(n, enemies))

Input

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

Output

True
raja
Published on 21-Oct-2020 14:10:05
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