Check if String Contains Consecutively Descending Substring in Python

Suppose we have a string s with some digits, we have to check whether it contains consecutively descending integers or not.

So, if the input is like s = "99989796", then the output will be True, as this string is holding [99,98,97,96]

To solve this, we will follow these steps−

• Define a function helper() . This will take pos, prev_num

• if pos is same as n, then

  • return True

• num_digits := digit count of prev_num

• for i in range num_digits - 1 to num_digits, do

  • if s[from index pos to pos+i-1] and numeric form of s[from index pos to pos+i-1]) is same as prev_num - 1, then

    • if helper(pos + i, prev_num - 1), then

    • return True

  • return False

  • From the main method, do the following−

  • n := size of s

  • for i in range 1 to quotient of n/2, do

    • num := numeric form of s[from index 0 to i-1]

    • if helper(i, num) is true, then

    • return True

  • return False

Let us see the following implementation to get better understanding −

Example

 Live Demo

class Solution:
def solve(self, s):
   n = len(s)
   def helper(pos, prev_num):
   if pos == n:
      return True
   num_digits = len(str(prev_num))
   for i in range(num_digits - 1, num_digits + 1):
      if s[pos:pos+i] and int(s[pos:pos+i]) == prev_num - 1:
         if helper(pos + i, prev_num - 1):
            return True
      return False
   for i in range(1, n//2 + 1):
      num = int(s[:i])
   if helper(i, num):
      return True
   return False
ob = Solution()
s = "99989796"
print(ob.solve(s))

Input

"99989796"

Output

True