Number of n digit stepping numbers

Given n, find count of n digit Stepping numbers. A number is called stepping number if all adjacent digits have an absolute difference of 1. 321 is a Stepping Number while 421 is not.

Examples :

Input : 2 
Output : 17
Explanation: The numbers are 10, 12, 21, 
23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 
78, 87, 89, 98.

Input : 1
Output : 10
Explanation: the numbers are 0, 1, 2, 3, 
4, 5, 6, 7, 8, 9.

A naive approach is to run a loop for all n digit numbers and check for every number if it is Stepping.

An efficient approach is to use dynamic programming.

In dp[i][j], i denotes number of
digits and j denotes last digit.

// If there is only one digit
if (i == 1)
   dp(i, j) = 1;

// If last digit is 0.
if (j == 0) 
   dp(i, j) = dp(i-1, j+1)

// If last digit is 9
else if (j == 9) 
   dp(i, j) = dp(i-1, j-1)

// If last digit is neither 0
// nor 9.
else 
   dp(i, j) = dp(i-1, j-1) + 
              dp(i-1, j+1)

Result is ∑dp(n, j) where j varies
from 1 to 9.

C++

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// CPP program to calculate the number of
// n digit stepping numbers.
#include <bits/stdc++.h>
using namespace std;
  
// function that calculates the answer
long long answer(int n)
{
    // dp[i][j] stores count of i digit
    // stepping numbers ending with digit
    // j.
    int dp[n + 1][10];
  
    // if n is 1 then answer will be 10.
    if (n == 1)
        return 10;
  
    // Initialize values for count of
    // digits equal to 1.
    for (int j = 0; j <= 9; j++)
        dp[1][j] = 1;
  
    // Compute values for count of digits
    // more than 1.
    for (int i = 2; i <= n; i++) {
        for (int j = 0; j <= 9; j++) {
  
            // If ending digit is 0
            if (j == 0)
                dp[i][j] = dp[i - 1][j + 1];
  
            // If ending digit is 9
            else if (j == 9)
                dp[i][j] = dp[i - 1][j - 1];
  
            // For other digits.
            else
                dp[i][j] = dp[i - 1][j - 1] + 
                           dp[i - 1][j + 1];
        }
    }
  
    // stores the final answer
    long long sum = 0;
    for (int j = 1; j <= 9; j++)
        sum += dp[n][j];
    return sum;
}
  
// driver program to test the above function
int main()
{
    int n = 2;
    cout << answer(n);
    return 0;
}

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Java














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// Java program to calculate the number of 


// n digit stepping numbers. 


  


class GFG { 


          


    // function that calculates the answer 


    static long answer(int n) 


    { 


        // dp[i][j] stores count of i  


        // digit stepping numbers ending  


        // with digit j. 


        int dp[][] = new int[n+1][10]; 


       


        // if n is 1 then answer will be 10. 


        if (n == 1) 


            return 10; 


       


        // Initialize values for count of 


        // digits equal to 1. 


        for (int j = 0; j <= 9; j++) 


            dp[1][j] = 1; 


       


        // Compute values for count of  


        // digits more than 1. 


        for (int i = 2; i <= n; i++) { 


            for (int j = 0; j <= 9; j++) { 


       


                // If ending digit is 0 


                if (j == 0) 


                    dp[i][j] = dp[i - 1][j + 1]; 


       


                // If ending digit is 9 


                else if (j == 9) 


                    dp[i][j] = dp[i - 1][j - 1]; 


       


                // For other digits. 


                else


                    dp[i][j] = dp[i - 1][j - 1] +  


                               dp[i - 1][j + 1]; 


            } 


        } 


       


        // stores the final answer 


        long sum = 0; 


        for (int j = 1; j <= 9; j++) 


            sum += dp[n][j]; 


        return sum; 


    } 


       


    // driver program to test the above function 


    public static void main(String args[]) 


    { 


        int n = 2; 


        System.out.println(answer(n)); 


    } 


} 


  


/*This code is contributed by Nikita tiwari.*/



















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Python3














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# Python3 program to calculate  


# the number of n digit 


# stepping numbers. 


  


# function that calculates 


# the answer 


def answer(n): 


      


    # dp[i][j] stores count of  


    # i digit stepping numbers  


    # ending with digit j. 


    dp = [[0 for x in range(10)]  


             for y in range(n + 1)]; 


  


    # if n is 1 then answer 


    # will be 10. 


    if (n == 1): 


        return 10; 


    for j in range(10): 


        dp[1][j] = 1; 


  


    # Compute values for count  


    # of digits more than 1. 


    for i in range(2, n + 1):  


        for j in range(10):  


              


            # If ending digit is 0 


            if (j == 0): 


                dp[i][j] = dp[i - 1][j + 1]; 


                  


            # If ending digit is 9 


            elif (j == 9): 


                dp[i][j] = dp[i - 1][j - 1]; 


                  


            # For other digits. 


            else: 


                dp[i][j] = (dp[i - 1][j - 1] + 


                            dp[i - 1][j + 1]); 


                  


    # stores the final answer 


    sum = 0; 


    for j in range(1, 10): 


        sum = sum + dp[n][j]; 


    return sum; 


  


# Driver Code 


n = 2; 


print(answer(n)); 


  


# This code is contributed  


# by mits 



















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C#














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// C# program to calculate the number of 


// n digit stepping numbers. 


using System; 


  


class GFG { 


          


    // function that calculates the answer 


    static long answer(int n) 


    { 


          


        // dp[i][j] stores count of i  


        // digit stepping numbers ending  


        // with digit j. 


        int [,]dp = new int[n+1,10]; 


      


        // if n is 1 then answer will be 10. 


        if (n == 1) 


            return 10; 


      


        // Initialize values for count of 


        // digits equal to 1. 


        for (int j = 0; j <= 9; j++) 


            dp[1,j] = 1; 


      


        // Compute values for count of  


        // digits more than 1. 


        for (int i = 2; i <= n; i++) { 


            for (int j = 0; j <= 9; j++) { 


      


                // If ending digit is 0 


                if (j == 0) 


                    dp[i,j] = dp[i - 1,j + 1]; 


      


                // If ending digit is 9 


                else if (j == 9) 


                    dp[i,j] = dp[i - 1,j - 1]; 


      


                // For other digits. 


                else


                    dp[i,j] = dp[i - 1,j - 1] +  


                               dp[i - 1,j + 1]; 


            } 


        } 


      


        // stores the final answer 


        long sum = 0; 


        for (int j = 1; j <= 9; j++) 


            sum += dp[n,j]; 


              


        return sum; 


    } 


      


    // driver program to test the above function 


    public static void Main() 


    { 


        int n = 2; 


          


        Console.WriteLine(answer(n)); 


    } 


} 


  


/*This code is contributed by vt_m.*/



















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PHP














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<?php 


// PHP program to calculate  


// the number of n digit 


// stepping numbers. 


  


// function that calculates 


// the answer 


function answer($n) 


{ 


    // dp[i][j] stores count of  


    // i digit stepping numbers  


    // ending with digit j. 


  


    // if n is 1 then answer 


    // will be 10. 


    if ($n == 1) 


        return 10; 


    for ( $j = 0; $j <= 9; $j++) 


        $dp[1][$j] = 1; 


  


    // Compute values for count  


    // of digits more than 1. 


    for ($i = 2; $i <= $n; $i++)  


    { 


        for ($j = 0; $j <= 9; $j++)  


        { 


  


            // If ending digit is 0 


            if ($j == 0) 


                $dp[$i][$j] = $dp[$i - 1][$j + 1]; 


  


            // If ending digit is 9 


            else if ($j == 9) 


                $dp[$i][$j] = $dp[$i - 1][$j - 1]; 


  


            // For other digits. 


            else


                $dp[$i][$j] = $dp[$i - 1][$j - 1] +  


                               $dp[$i - 1][$j + 1]; 


        } 


    } 


  


    // stores the final answer 


    $sum = 0; 


    for ($j = 1; $j <= 9; $j++) 


        $sum += $dp[$n][$j]; 


    return $sum; 


} 


  


// Driver Code 


$n = 2; 


echo answer($n); 


  


// This code is contributed by aj_36 


?> 



















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Output :


17



Time Complexity: O(n)

Auxiliary Space: O(n)Number of n digit stepping numbers | Space optimized solution



          
          
          
          

            

    

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