Syntax error Program to count number of permutations where sum of adjacent pairs are perfect square in Python

Program to count number of permutations where sum of adjacent pairs are perfect square in Python

Suppose we have a list of numbers called nums. We have to find the number of permutations of nums such that sum of every pair of adjacent values is a perfect square. Two permutations A and B are unique when there is some index i where A[i] is not same as B[i].

So, if the input is like nums = [2, 9, 7], then the output will be 2, as we have [2, 7, 9] and [9, 7, 2]

To solve this, we will follow these steps −

  • res := 0

  • Define a function util() . This will take i

  • if i + 1 is same as size of nums , then

    • res := res + 1

    • return

  • visited := a new empty set

  • for j in range i + 1 to size of nums, do

    • s := nums[i] + nums[j]

    • if s is not visited and (square root of s)^2 is s, then

      • mark s as visited

      • swap nums[i + 1] and nums[j]

      • util(i + 1)

      • swap nums[i + 1] and nums[j]

  • From the main method do the following −

  • visited := a new set

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

    • swap nums[i] and nums[0]

    • if nums[0] is not visited, then

      • util(0)

    • mark nums[0] as visited

    • swap nums[i] and nums[0]

  • return res

Let us see the following implementation to get better understanding −

Example

 Live Demo

from math import sqrt
class Solution:
   def solve(self, nums):
      self.res = 0
      def util(i):
         if i + 1 == len(nums):
            self.res += 1
            return
         visited = set()
         for j in range(i + 1, len(nums)):
            s = nums[i] + nums[j]
            if s not in visited and int(sqrt(s)) ** 2 == s:
               visited.add(s)
               nums[i + 1], nums[j] = nums[j], nums[i + 1]
               util(i + 1)
               nums[i + 1], nums[j] = nums[j], nums[i + 1]
      visited = set()
      for i in range(len(nums)):
         nums[i], nums[0] = nums[0], nums[i]
         if nums[0] not in visited:
            util(0)
         visited.add(nums[0])
         nums[i], nums[0] = nums[0], nums[i]
      return self.res
ob = Solution()
nums = [2, 9, 7]
print(ob.solve(nums))

Input

[2, 9, 7]

Output

2
Updated on: 2020-12-26T10:40:54+05:30

232 Views

Kickstart Your Career

Get certified by completing the course

Get Started
Advertisements