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Print Nth Stepping or Autobiographical number
  • Last Updated : 05 May, 2020

Given a natural number N, the task is to print the Nth Stepping or Autobiographical number.

A number is called stepping number if all adjacent digits have an absolute difference of 1. The following series is a list of Stepping natural numbers:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 21, 22, 23, 32, ….

Examples:

Input: N = 16 
Output: 32
Explanation:
16th Stepping number is 32.

Input: N = 14 
Output: 22
Explanation:
14th Stepping number is 22.

Approach: This problem can be solved using Queue data structure. First, prepare an empty queue, and Enqueue 1, 2, …, 9 in this order.
Then inorder the generate the Nth Stepping number, the following operations has to be performed N times:

  • Perform Dequeue from the Queue. Let x be the dequeued element.
  • If x mod 10 is not equal to 0, then Enqueue 10x + (x mod 10) – 1
  • Enqueue 10x + (x mod 10).
  • If x mod 10 is not equal to 9, then Enqueue 10x + (x mod 10) + 1.

The dequeued number in the N-th operation is the N-th Stepping Number.



Below is the implementation of the above approach:

C++




// C++ implementation to find
// N’th stepping natural Number
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the
// Nth stepping natural number
int NthSmallest(int K)
{
  
    // Declare the queue
    queue<int> Q;
  
    int x;
  
    // Enqueue 1, 2, ..., 9 in this order
    for (int i = 1; i < 10; i++)
        Q.push(i);
  
    // Perform K operation on queue
    for (int i = 1; i <= K; i++) {
  
        // Get the ith Stepping number
        x = Q.front();
  
        // Perform Dequeue from the Queue
        Q.pop();
  
        // If x mod 10 is not equal to 0
        if (x % 10 != 0) {
  
            // then Enqueue 10x + (x mod 10) - 1
            Q.push(x * 10 + x % 10 - 1);
        }
  
        // Enqueue 10x + (x mod 10)
        Q.push(x * 10 + x % 10);
  
        // If x mod 10 is not equal to 9
        if (x % 10 != 9) {
  
            // then Enqueue 10x + (x mod 10) + 1
            Q.push(x * 10 + x % 10 + 1);
        }
    }
  
    // Return the dequeued number of the K-th
    // operation as the Nth stepping number
    return x;
}
  
// Driver Code
int main()
{
  
    // initialise K
    int N = 16;
  
    cout << NthSmallest(N) << "\n";
  
    return 0;
}

Java




// Java implementation to find
// N'th stepping natural Number
import java.util.*;
  
class GFG{
   
// Function to find the
// Nth stepping natural number
static int NthSmallest(int K)
{
   
    // Declare the queue
    Queue<Integer> Q = new LinkedList<>();
   
    int x = 0;
   
    // Enqueue 1, 2, ..., 9 in this order
    for (int i = 1; i < 10; i++)
        Q.add(i);
   
    // Perform K operation on queue
    for (int i = 1; i <= K; i++) {
   
        // Get the ith Stepping number
        x = Q.peek();
   
        // Perform Dequeue from the Queue
        Q.remove();
   
        // If x mod 10 is not equal to 0
        if (x % 10 != 0) {
   
            // then Enqueue 10x + (x mod 10) - 1
            Q.add(x * 10 + x % 10 - 1);
        }
   
        // Enqueue 10x + (x mod 10)
        Q.add(x * 10 + x % 10);
   
        // If x mod 10 is not equal to 9
        if (x % 10 != 9) {
   
            // then Enqueue 10x + (x mod 10) + 1
            Q.add(x * 10 + x % 10 + 1);
        }
    }
   
    // Return the dequeued number of the K-th
    // operation as the Nth stepping number
    return x;
}
   
// Driver Code
public static void main(String[] args)
{
   
    // initialise K
    int N = 16;
   
    System.out.print(NthSmallest(N));
}
}
  
// This code is contributed by 29AjayKumar

Python3




# Python3 implementation to find
# N’th stepping natural Number
  
# Function to find the
# Nth stepping natural number
def NthSmallest(K):
    # Declare the queue
    Q = []
  
    # Enqueue 1, 2, ..., 9 in this order
    for i in range(1,10):
        Q.append(i)
  
    # Perform K operation on queue
    for i in range(1,K+1):
        # Get the ith Stepping number
        x = Q[0]
  
        # Perform Dequeue from the Queue
        Q.remove(Q[0])
  
        # If x mod 10 is not equal to 0
        if (x % 10 != 0):
            # then Enqueue 10x + (x mod 10) - 1
            Q.append(x * 10 + x % 10 - 1)
  
        # Enqueue 10x + (x mod 10)
        Q.append(x * 10 + x % 10)
  
        # If x mod 10 is not equal to 9
        if (x % 10 != 9):
            # then Enqueue 10x + (x mod 10) + 1
            Q.append(x * 10 + x % 10 + 1)
  
    # Return the dequeued number of the K-th
    # operation as the Nth stepping number
    return x
  
# Driver Code
if __name__ == '__main__':
    # initialise K
    N = 16
  
    print(NthSmallest(N))
  
# This code is contributed by Surendra_Gangwar

C#




// C# implementation to find
// N'th stepping natural Number
using System;
using System.Collections.Generic;
  
class GFG{
    
// Function to find the
// Nth stepping natural number
static int NthSmallest(int K)
{
    
    // Declare the queue
    List<int> Q = new List<int>();
    
    int x = 0;
    
    // Enqueue 1, 2, ..., 9 in this order
    for (int i = 1; i < 10; i++)
        Q.Add(i);
    
    // Perform K operation on queue
    for (int i = 1; i <= K; i++) {
    
        // Get the ith Stepping number
        x = Q[0];
    
        // Perform Dequeue from the Queue
        Q.RemoveAt(0);
    
        // If x mod 10 is not equal to 0
        if (x % 10 != 0) {
    
            // then Enqueue 10x + (x mod 10) - 1
            Q.Add(x * 10 + x % 10 - 1);
        }
    
        // Enqueue 10x + (x mod 10)
        Q.Add(x * 10 + x % 10);
    
        // If x mod 10 is not equal to 9
        if (x % 10 != 9) {
    
            // then Enqueue 10x + (x mod 10) + 1
            Q.Add(x * 10 + x % 10 + 1);
        }
    }
    
    // Return the dequeued number of the K-th
    // operation as the Nth stepping number
    return x;
}
    
// Driver Code
public static void Main(String[] args)
{
    
    // initialise K
    int N = 16;
    
    Console.Write(NthSmallest(N));
}
}
  
// This code is contributed by sapnasingh4991
Output:
32

Time Complexity: O(N)

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