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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 &Sum;dp(n, j) where j varies
from 1 to 9.

C++

// 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; 
} 

Java

// 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.*/

Python3

# 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 

C#

// 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.*/

PHP

<?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 
?> 