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++
#include <bits/stdc++.h>
using namespace std;
long long answer( int n)
{
int dp[n + 1][10];
if (n == 1)
return 10;
for ( int j = 0; j <= 9; j++)
dp[1][j] = 1;
for ( int i = 2; i <= n; i++) {
for ( int j = 0; j <= 9; j++) {
if (j == 0)
dp[i][j] = dp[i - 1][j + 1];
else if (j == 9)
dp[i][j] = dp[i - 1][j - 1];
else
dp[i][j] = dp[i - 1][j - 1] +
dp[i - 1][j + 1];
}
}
long long sum = 0;
for ( int j = 1; j <= 9; j++)
sum += dp[n][j];
return sum;
}
int main()
{
int n = 2;
cout << answer(n);
return 0;
}
|
Java
class GFG {
static long answer( int n)
{
int dp[][] = new int [n+ 1 ][ 10 ];
if (n == 1 )
return 10 ;
for ( int j = 0 ; j <= 9 ; j++)
dp[ 1 ][j] = 1 ;
for ( int i = 2 ; i <= n; i++) {
for ( int j = 0 ; j <= 9 ; j++) {
if (j == 0 )
dp[i][j] = dp[i - 1 ][j + 1 ];
else if (j == 9 )
dp[i][j] = dp[i - 1 ][j - 1 ];
else
dp[i][j] = dp[i - 1 ][j - 1 ] +
dp[i - 1 ][j + 1 ];
}
}
long sum = 0 ;
for ( int j = 1 ; j <= 9 ; j++)
sum += dp[n][j];
return sum;
}
public static void main(String args[])
{
int n = 2 ;
System.out.println(answer(n));
}
}
|
Python3
def answer(n):
dp = [[ 0 for x in range ( 10 )]
for y in range (n + 1 )];
if (n = = 1 ):
return 10 ;
for j in range ( 10 ):
dp[ 1 ][j] = 1 ;
for i in range ( 2 , n + 1 ):
for j in range ( 10 ):
if (j = = 0 ):
dp[i][j] = dp[i - 1 ][j + 1 ];
elif (j = = 9 ):
dp[i][j] = dp[i - 1 ][j - 1 ];
else :
dp[i][j] = (dp[i - 1 ][j - 1 ] +
dp[i - 1 ][j + 1 ]);
sum = 0 ;
for j in range ( 1 , 10 ):
sum = sum + dp[n][j];
return sum ;
n = 2 ;
print (answer(n));
|
C#
using System;
class GFG {
static long answer( int n)
{
int [,]dp = new int [n+1,10];
if (n == 1)
return 10;
for ( int j = 0; j <= 9; j++)
dp[1,j] = 1;
for ( int i = 2; i <= n; i++) {
for ( int j = 0; j <= 9; j++) {
if (j == 0)
dp[i,j] = dp[i - 1,j + 1];
else if (j == 9)
dp[i,j] = dp[i - 1,j - 1];
else
dp[i,j] = dp[i - 1,j - 1] +
dp[i - 1,j + 1];
}
}
long sum = 0;
for ( int j = 1; j <= 9; j++)
sum += dp[n,j];
return sum;
}
public static void Main()
{
int n = 2;
Console.WriteLine(answer(n));
}
}
|
PHP
<?php
function answer( $n )
{
if ( $n == 1)
return 10;
for ( $j = 0; $j <= 9; $j ++)
$dp [1][ $j ] = 1;
for ( $i = 2; $i <= $n ; $i ++)
{
for ( $j = 0; $j <= 9; $j ++)
{
if ( $j == 0)
$dp [ $i ][ $j ] = $dp [ $i - 1][ $j + 1];
else if ( $j == 9)
$dp [ $i ][ $j ] = $dp [ $i - 1][ $j - 1];
else
$dp [ $i ][ $j ] = $dp [ $i - 1][ $j - 1] +
$dp [ $i - 1][ $j + 1];
}
}
$sum = 0;
for ( $j = 1; $j <= 9; $j ++)
$sum += $dp [ $n ][ $j ];
return $sum ;
}
$n = 2;
echo answer( $n );
?>
|
Javascript
<script>
function answer(n)
{
let dp = new Array(n + 1);
for ( var i = 0; i < dp.length; i++)
{
dp[i] = new Array(2);
}
if (n == 1)
return 10;
for (let j = 0; j <= 9; j++)
dp[1][j] = 1;
for (let i = 2; i <= n; i++)
{
for (let j = 0; j <= 9; j++)
{
if (j == 0)
dp[i][j] = dp[i - 1][j + 1];
else if (j == 9)
dp[i][j] = dp[i - 1][j - 1];
else
dp[i][j] = dp[i - 1][j - 1] +
dp[i - 1][j + 1];
}
}
let sum = 0;
for (let j = 1; j <= 9; j++)
sum += dp[n][j];
return sum;
}
let n = 2;
document.write(answer(n));
</script>
|
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++
#include <bits/stdc++.h>
using namespace std;
long long answer( int n)
{
vector< long long > dp(10, 1);
if (n == 1)
return 10;
for ( int i = 2; i <= n; i++) {
vector< long long > new_dp(10, 0);
new_dp[0] = dp[1];
new_dp[9] = dp[8];
for ( int j = 1; j <= 8; j++)
new_dp[j] = dp[j - 1] + dp[j + 1];
dp = new_dp;
}
long long sum = 0;
for ( int j = 1; j <= 9; j++)
sum += dp[j];
return sum;
}
int main()
{
int n = 2;
cout << answer(n);
return 0;
}
|
Java
import java.util.*;
public class Main
{
static long answer( int n)
{
ArrayList<Long> dp = new ArrayList<>(Collections.nCopies( 10 , 1L));
if (n == 1 )
return 10 ;
for ( int i = 2 ; i <= n; i++) {
ArrayList<Long> new_dp = new ArrayList<>(Collections.nCopies( 10 , 0L));
new_dp.set( 0 , dp.get( 1 ));
new_dp.set( 9 , dp.get( 8 ));
for ( int j = 1 ; j <= 8 ; j++)
new_dp.set(j, dp.get(j - 1 ) + dp.get(j + 1 ));
dp = new_dp;
}
long sum = 0 ;
for ( int j = 1 ; j <= 9 ; j++)
sum += dp.get(j);
return sum;
}
public static void main(String[] args) {
int n = 2 ;
System.out.println(answer(n));
}
}
|
Python3
def answer(n):
dp = [ 1 ] * 10
if n = = 1 :
return 10
for i in range ( 2 , n + 1 ):
new_dp = [ 0 ] * 10
new_dp[ 0 ] = dp[ 1 ]
new_dp[ 9 ] = dp[ 8 ]
for j in range ( 1 , 9 ):
new_dp[j] = dp[j - 1 ] + dp[j + 1 ]
dp = new_dp
sum = 0
for j in range ( 1 , 10 ):
sum + = dp[j]
return sum
n = 2
print (answer(n))
|
C#
using System;
using System.Collections.Generic;
class SteppingNumbers {
static long Answer( int n)
{
List< long > dp = new List< long >();
for ( int i = 0; i < 10; i++) {
dp.Add(1);
}
if (n == 1) {
return 10;
}
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);
}
new_dp[0] = dp[1];
new_dp[9] = dp[8];
for ( int j = 1; j <= 8; j++) {
new_dp[j] = dp[j - 1] + dp[j + 1];
}
dp = new_dp;
}
long sum = 0;
for ( int j = 1; j <= 9; j++) {
sum += dp[j];
}
return sum;
}
static void Main()
{
int n = 2;
Console.WriteLine(Answer(n));
}
}
|
Javascript
function answer(n) {
let dp = new Array(10).fill(1);
if (n == 1)
return 10;
for (let i = 2; i <= n; i++) {
let new_dp = new Array(10).fill(0);
new_dp[0] = dp[1];
new_dp[9] = dp[8];
for (let j = 1; j <= 8; j++)
new_dp[j] = dp[j - 1] + dp[j + 1];
dp = new_dp;
}
let sum = 0;
for (let j = 1; j <= 9; j++)
sum += dp[j];
return sum;
}
let n = 2;
console.log(answer(n));
|
Time complexity: O(n)
Auxiliary Space: O(10) => O(1)
Number of n digit stepping numbers | Space optimized solution
Last Updated :
24 Apr, 2023
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