Find All Upside-Down Numbers of Length n in Python

Suppose we have a value n. We have to find all upside down numbers of length n. As we knot the upside down number is one that appears the same when flipped 180 degrees.

So, if the input is like n = 2, then the output will be ['11', '69', '88', '96'].

To solve this, we will follow these steps −

• Define a function middle() . This will take x

• if x is 0, then

  • return list of a blank string

• if x is same as 1, then

  • return a new list of elements 0, 1, 8

• ret := a new list

• mid := middle(x − 2)

• for each m in mid, do

  • if x is not same as n, then

    • insert ("0" concatenate m concatenate "0") at the end of ret

  • insert ("1" concatenate m concatenate "1") at the end of ret

  • insert ("6" concatenate m concatenate "9") at the end of ret

  • insert ("8" concatenate m concatenate "8") at the end of ret

  • insert ("9" concatenate m concatenate "6") at the end of ret

• return ret

• From the main method, do the following −

• if n is 0, then

  • return a new list

• otherwise return sorted list of middle(n)

Let us see the following implementation to get better understanding −

Example

 Live Demo

class Solution:
   def solve(self, n):
      if not n:
         return []
      def middle(x=n):
         if not x:
            return [""]
         if x == 1:
            return list("018")
         ret = []
         mid = middle(x - 2)
         for m in mid:
            if x != n:
               ret.append("0" + m + "0")
            ret.append("1" + m + "1")
            ret.append("6" + m + "9")
            ret.append("8" + m + "8")
            ret.append("9" + m + "6")
         return ret
      return sorted(middle())
ob = Solution()
print(ob.solve(2))

Input

2

Output

['11', '69', '88', '96']