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++

// 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
?>

Javascript

<script>

// JavaScript 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.
    let dp = new Array(n + 1);

    // Loop to create 2D array using 1D array
    for(var i = 0; i < dp.length; i++)
    {
        dp[i] = new Array(2);
    }

    // If n is 1 then answer will be 10.
    if (n == 1)
        return 10;

    // Initialize values for count of
    // digits equal to 1.
    for(let j = 0; j <= 9; j++)
        dp[1][j] = 1;

    // Compute values for count of
    // digits more than 1.
    for(let i = 2; i <= n; i++)
    {
        for(let 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
    let sum = 0;
    for(let j = 1; j <= 9; j++)
        sum += dp[n][j];

    return sum;
}

// Driver Code
let n = 2;

document.write(answer(n));

// This code is contributed by code_hunt

</script>

Output

17

Time Complexity: O(n), as we are using a loop to traverse n times and within each iteration there are 10 cases. So a total of 10*N that is equivalent to n.
Auxiliary Space: O(n), as we are using extra space of 10*n for dp array.

Efficient approach: Space optimization

In the previous approach, the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.

Implementation steps:

• Create a 1D vector dp of size 10 and initialize it with 1.
• Set a base case.
• Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
• Create a variable sum and and update it by iterating over DP. 
• At last return and print the final answer stored in sum .

Implementation:

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] stores count of i digit
    // stepping numbers ending with digits
    // 0 to 9.
    vector<long long> dp(10, 1);

    // if n is 1 then answer will be 10.
    if (n == 1)
        return 10;

    // Compute values for count of digits
    // more than 1.
    for (int i = 2; i <= n; i++) {
        vector<long long> new_dp(10, 0);

        // If ending digit is 0
        new_dp[0] = dp[1];

        // If ending digit is 9
        new_dp[9] = dp[8];

        // For other digits.
        for (int j = 1; j <= 8; j++)
            new_dp[j] = dp[j - 1] + dp[j + 1];

        dp = new_dp;
    }

    // stores the final answer
    long long sum = 0;
    for (int j = 1; j <= 9; j++)
        sum += dp[j];
    return sum;
}

// driver program to test the above function
int main()
{
    int n = 2;
    cout << answer(n);
    return 0;
}

Java

import java.util.*;

public class Main
{

  // function that calculates the answer
  static long answer(int n)
  {

    // dp[i] stores count of i digit
    // stepping numbers ending with digits
    // 0 to 9.
    ArrayList<Long> dp = new ArrayList<>(Collections.nCopies(10, 1L));

    // if n is 1 then answer will be 10.
    if (n == 1)
      return 10;

    // Compute values for count of digits
    // more than 1.
    for (int i = 2; i <= n; i++) {
      ArrayList<Long> new_dp = new ArrayList<>(Collections.nCopies(10, 0L));

      // If ending digit is 0
      new_dp.set(0, dp.get(1));

      // If ending digit is 9
      new_dp.set(9, dp.get(8));

      // For other digits.
      for (int j = 1; j <= 8; j++)
        new_dp.set(j, dp.get(j - 1) + dp.get(j + 1));

      dp = new_dp;
    }

    // stores the final answer
    long sum = 0;
    for (int j = 1; j <= 9; j++)
      sum += dp.get(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 Prajwal Kandekar

Python3

def answer(n):
    # dp[i] stores count of i digit
    # stepping numbers ending with digits
    # 0 to 9.
    dp = [1] * 10

    # if n is 1 then answer will be 10.
    if n == 1:
        return 10

    # Compute values for count of digits
    # more than 1.
    for i in range(2, n + 1):
        new_dp = [0] * 10

        # If ending digit is 0
        new_dp[0] = dp[1]

        # If ending digit is 9
        new_dp[9] = dp[8]

        # For other digits.
        for j in range(1, 9):
            new_dp[j] = dp[j - 1] + dp[j + 1]

        dp = new_dp

    # stores the final answer
    sum = 0
    for j in range(1, 10):
        sum += dp[j]
    return sum

# driver program to test the above function
n = 2
print(answer(n))

C#

using System;
using System.Collections.Generic;

class SteppingNumbers {
    // function that calculates the answer
    static long Answer(int n)
    {
        // dp[i] stores count of i digit
        // stepping numbers ending with digits
        // 0 to 9.
        List<long> dp = new List<long>();
        for (int i = 0; i < 10; i++) {
            dp.Add(1);
        }

        // if n is 1 then answer will be 10.
        if (n == 1) {
            return 10;
        }

        // Compute values for count of digits
        // more than 1.
        for (int i = 2; i <= n; i++) {
            List<long> new_dp = new List<long>();
            for (int j = 0; j < 10; j++) {
                new_dp.Add(0);
            }

            // If ending digit is 0
            new_dp[0] = dp[1];

            // If ending digit is 9
            new_dp[9] = dp[8];

            // For other digits.
            for (int j = 1; j <= 8; j++) {
                new_dp[j] = dp[j - 1] + dp[j + 1];
            }

            dp = new_dp;
        }

        // stores the final answer
        long sum = 0;
        for (int j = 1; j <= 9; j++) {
            sum += dp[j];
        }
        return sum;
    }

    // driver program to test the above function
    static void Main()
    {
        int n = 2;
        Console.WriteLine(Answer(n));
    }
}

Javascript

function answer(n) {

    // dp[i] stores count of i digit
    // stepping numbers ending with digits
    // 0 to 9.
    let dp = new Array(10).fill(1);

    // if n is 1 then answer will be 10.
    if (n == 1)
        return 10;

    // Compute values for count of digits
    // more than 1.
    for (let i = 2; i <= n; i++) {
        let new_dp = new Array(10).fill(0);

        // If ending digit is 0
        new_dp[0] = dp[1];

        // If ending digit is 9
        new_dp[9] = dp[8];

        // For other digits.
        for (let j = 1; j <= 8; j++)
            new_dp[j] = dp[j - 1] + dp[j + 1];

        dp = new_dp;
    }

    // stores the final answer
    let sum = 0;
    for (let j = 1; j <= 9; j++)
        sum += dp[j];
    return sum;
}

// driver program to test the above function
let n = 2;
console.log(answer(n));

Output

17

Time complexity: O(n)
Auxiliary Space: O(10) => O(1)

Number of n digit stepping numbers | Space optimized solution