Given n, find the count of n digit Stepping numbers. A number is called a 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 The numbers are 10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87, 89, 98. Input : 1 Output : 10 The numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
In the previous post, a solution that requires O(n) auxiliary space is discussed. The auxiliary space required to solve the problem can be optimized. The 2-D dp array dp[i][j] represents count of stepping number of length i and last digit j. For a digit j the count is obtained from digit j – 1 and j + 1. The recurrence relation is dp[i][j] = dp[i-1][j-1] + dp[i-1][j+1] . Observe that the answer for current length i depends only on i – 1. So a 1-D dp array can be used in which for a given i, dp[j] stores count of stepping numbers of length i ending with digit j. Before updating dp array for a given length i, store the result for length i – 1 in another array prev, then update dp array using prev array.
Below is the implementation of above approach:
C++
// CPP program to calculate the number of// n digit stepping numbers.#include <bits/stdc++.h>using namespace std;// function that calculates the answerlong long answer(int n){ // dp[j] stores count of i digit // stepping numbers ending with digit // j. int dp[10]; // To store result of length i - 1 // before updating dp[j] for length i. int prev[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[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++) { prev[j] = dp[j]; } for (int j = 0; j <= 9; j++) { // If ending digit is 0 if (j == 0) dp[j] = prev[j + 1]; // If ending digit is 9 else if (j == 9) dp[j] = prev[j - 1]; // For other digits. else dp[j] = prev[j - 1] + prev[j + 1]; } } // 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 functionint 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 answerstatic long answer(int n){ // dp[j] stores count of i digit // stepping numbers ending with digit // j. int[] dp = new int[10]; // To store result of length i - 1 // before updating dp[j] for length i. int[] prev = new int[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[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++) { prev[j] = dp[j]; } for (int j = 0; j <= 9; j++) { // If ending digit is 0 if (j == 0) dp[j] = prev[j + 1]; // If ending digit is 9 else if (j == 9) dp[j] = prev[j - 1]; // For other digits. else dp[j] = prev[j - 1] + prev[j + 1]; } } // stores the final answer long sum = 0; for (int j = 1; j <= 9; j++) sum += dp[j]; return sum;}// Driver codepublic static void main (String[] args) { int n = 2; System.out.println(answer(n));}}// This code is contributed by mits |
Python3
# Python3 program to calculate the number of# n digit stepping numbers.# function that calculates the answerdef answer(n) : # dp[j] stores count of i digit # stepping numbers ending with digit j. dp = [0] * 10 # To store resu1lt of length i - 1 # before updating dp[j] for length i. prev = [0] * 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 j in range(0, 10) : dp[j] = 1 # Compute values for count of digits # more than 1. for i in range(2, n + 1): for j in range (0, 10): prev[j] = dp[j] for j in range (0, 10): # If ending digit is 0 if (j == 0): dp[j] = prev[j + 1] # If ending digit is 9 elif (j == 9) : dp[j] = prev[j - 1] # For other digits. else : dp[j] = prev[j - 1] + prev[j + 1] # stores the final answer sum = 0 for j in range (1, 10): sum = sum + dp[j] return sum# Driver Coden = 2print(answer(n))# This code is contributed by ihritik |
C#
// C# program to calculate the number of// n digit stepping numbers.using System;class GFG{ // function that calculates the answerstatic long answer(int n){ // dp[j] stores count of i digit // stepping numbers ending with digit // j. int[] dp = new int[10]; // To store result of length i - 1 // before updating dp[j] for length i. int[] prev = new int[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[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++) { prev[j] = dp[j]; } for (int j = 0; j <= 9; j++) { // If ending digit is 0 if (j == 0) dp[j] = prev[j + 1]; // If ending digit is 9 else if (j == 9) dp[j] = prev[j - 1]; // For other digits. else dp[j] = prev[j - 1] + prev[j + 1]; } } // stores the final answer long sum = 0; for (int j = 1; j <= 9; j++) sum += dp[j]; return sum;}// Driver codestatic void Main(){ int n = 2; Console.WriteLine(answer(n));}}// This code is contributed by mits |
PHP
<?php// PHP program to calculate the number of// n digit stepping numbers.// function that calculates the answerfunction answer($n){ // dp[j] stores count of i digit // stepping numbers ending with digit // j. $dp = array_fill(0, 10, 0); // To store result of length i - 1 // before updating dp[j] for length i. $prev = array_fill(0, 10, 0);; // if n is 1 then answer will be 10. if ($n == 1) return 10; // Initialize values for count of // digits equal to 1. for ($j = 0; $j <= 9; $j++) $dp[$j] = 1; // Compute values for count of digits // more than 1. for ($i = 2; $i <= $n; $i++) { for ($j = 0; $j <= 9; $j++) { $prev[$j] = $dp[$j]; } for ($j = 0; $j <= 9; $j++) { // If ending digit is 0 if ($j == 0) $dp[$j] = $prev[$j + 1]; // If ending digit is 9 else if ($j == 9) $dp[$j] = $prev[$j - 1]; // For other digits. else $dp[$j] = $prev[$j - 1] + $prev[$j + 1]; } } // stores the final answer $sum = 0; for ($j = 1; $j <= 9; $j++) $sum += $dp[$j]; return $sum;} // Driver program to test the above function $n = 2; echo answer($n);// This code is contributed by mits?> |
Javascript
<script>// Javascript program to calculate the number of// n digit stepping numbers.// function that calculates the answerfunction answer(n){ // dp[j] stores count of i digit // stepping numbers ending with digit // j. var dp = Array(10); // To store result of length i - 1 // before updating dp[j] for length i. var prev = Array(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 (var j = 0; j <= 9; j++) dp[j] = 1; // Compute values for count of digits // more than 1. for (var i = 2; i <= n; i++) { for (var j = 0; j <= 9; j++) { prev[j] = dp[j]; } for (var j = 0; j <= 9; j++) { // If ending digit is 0 if (j == 0) dp[j] = prev[j + 1]; // If ending digit is 9 else if (j == 9) dp[j] = prev[j - 1]; // For other digits. else dp[j] = prev[j - 1] + prev[j + 1]; } } // stores the final answer var sum = 0; for (var j = 1; j <= 9; j++) sum += dp[j]; return sum;}// driver program to test the above functionvar n = 2;document.write( answer(n));</script> |
Output:
17
Time Complexity: O(N)
Auxiliary Space: O(1), since no extra space has been taken.
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