Program to count number of stepping numbers of n digits in python

Suppose we have a number n we have to find the number of stepping numbers of n digit. As we know a number is called stepping number when all adjacent digits have an absolute difference of 1. So if a number is 123, this is stepping number but 124 is not. If the answer is very large then return result mod 10^9 + 7.

So, if the input is like n = 2, then the output will be 17, as the stepping numbers are [12, 23, 34, 45, 56, 67, 78, 89, 98, 87, 76, 65, 54, 43, 32, 21, 10]

To solve this, we will follow these steps −

• m := 10^9 + 7
• if n is same as 0, then
  • return 0
• if n is same as 1, then
  • return 10
• dp := a list of size 10 filled with value 1
• iterate n - 1 times, do
  • ndp := a list of size 10 filled with value 0
  • ndp[0] := dp[1]
  • for i in range 1 to 8, do
    • ndp[i] := dp[i - 1] + dp[i + 1]
  • ndp[9] := dp[8]
  • dp := ndp
• return (sum of all numbers of dp[from index 1 to end]) mod m

Let us see the following implementation to get better understanding −

Example 

Live Demo

class Solution:
   def solve(self, n):
      m = (10 ** 9 + 7)
      if n == 0:
         return 0
      if n == 1:
         return 10
      dp = [1] * 10
      for _ in range(n - 1):
         ndp = [0] * 10
         ndp[0] = dp[1]
         for i in range(1, 9):
            ndp[i] = dp[i - 1] + dp[i + 1]
         ndp[9] = dp[8]
         dp = ndp
      return sum(dp[1:]) % m

ob = Solution()
n = 3
print(ob.solve(n))

Input

3

Output

32

Arnab Chakraborty

Updated on: 02-Dec-2020

368 Views

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