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 ?> |
Output :
17
Time Complexity: O(n)
Auxiliary Space: O(n)Number of n digit stepping numbers | Space optimized solution
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