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
?>

Output :

Time Complexity: O(n)
Auxiliary Space: O(n)

My Personal Notes

Striver

I am weird I am weirdo u think u know me Nah u dont know me https//wwwfacebookcom/rajstriver

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Improved By : jit_t, Mithun Kumar

Practice Tags :

Dynamic Programming

Mathematical