Nth positive number whose absolute difference of adjacent digits is at most 1


Given a number N, the task is to find the Nth number which has an absolute difference of 1 between every pair of its adjacent digits.

Examples:

Input : N = 5
Output : 5
Explanation:
The first 5 such numbers are 1,2,3,4 and 5.

Input : N = 15
Output : 23
Explanation:
The first 15 such numbers are 1,2,3,4,5,6,7,8,9,10,11,12,21,22 and 23.

Approach: In order to solve this problem we are using the Queue data structure.



  • Prepare an empty Queue, and Enqueue all integers 1 to 9 in increasing order.
  • Now perform the following operation N times.
    • Dequeue and store in array arr which stores ith number of required type in arr[i].
    • If (arr[i] % 10 != 0), then enqueue 10 * arr[i] + (arr[i] % 10) – 1.
    • Enqueue 10 * arr[i] + (arr[i] % 10).
    • If (arr[i] % 10 != 9), then enqueue 10 * arr[i] + (arr[i] % 10) + 1.
  • Return arr[N] as the answer.

Below is the implementation of the given approach:

C++

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// C++ Program to find Nth number with
// absolute difference between all 
// adjacent digits at most 1.
  
#include <bits/stdc++.h>
using namespace std;
  
  
// Return Nth number with
// absolute difference between all 
// adjacent digits at most 1.
void findNthNumber(int N)
{
    // To store all such numbers
    long long arr[N + 1];
      
    queue<long long> q;
  
    // Enqueue all integers from 1 to 9 
    // in increasing order.
    for (int i = 1; i <= 9; i++)
        q.push(i);
  
    // Perform the operation N times so that
    // we can get all such N numbers.
    for (int i = 1; i <= N; i++) {
  
        // Store the front element of queue,
        // in array and pop it from queue.
        arr[i] = q.front();
        q.pop();
  
        // If the last digit of dequeued integer is
        // not 0, then enqueue the next such number.
        if (arr[i] % 10 != 0)
            q.push(arr[i] * 10 + arr[i] % 10 - 1);
  
        // Enqueue the next such number
        q.push(arr[i] * 10 + arr[i] % 10);
  
        // If the last digit of dequeued integer is
        // not 9, then enqueue the next such number.
        if (arr[i] % 10 != 9)
            q.push(arr[i] * 10 + arr[i] % 10 + 1);
    }
      
    cout<<arr[N]<<endl;
}
  
// Driver Code
int main()
{
    int N = 21;
    findNthNumber(N);
    return 0;
}

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Java

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// Java program to find Nth number with
// absolute difference between all 
// adjacent digits at most 1.
import java.util.*;
  
class GFG{
  
// Return Nth number with
// absolute difference between all 
// adjacent digits at most 1.
static void findNthNumber(int N)
{
      
    // To store all such numbers
    int []arr = new int[N + 1];
      
    Queue<Integer> q = new LinkedList<>();
  
    // Enqueue all integers from 1 to 9 
    // in increasing order.
    for(int i = 1; i <= 9; i++)
       q.add(i);
  
    // Perform the operation N times so 
    // that we can get all such N numbers.
    for(int i = 1; i <= N; i++)
    {
         
       // Store the front element of queue,
       // in array and pop it from queue.
       arr[i] = q.peek();
       q.remove();
         
       // If the last digit of dequeued 
       // integer is not 0, then enqueue
       // the next such number.
       if (arr[i] % 10 != 0)
           q.add(arr[i] * 10 + arr[i] % 10 - 1);
         
       // Enqueue the next such number
       q.add(arr[i] * 10 + arr[i] % 10);
         
       // If the last digit of dequeued 
       // integer is not 9, then enqueue 
       // the next such number.
       if (arr[i] % 10 != 9)
           q.add(arr[i] * 10 + arr[i] % 10 + 1);
    }
    System.out.println(arr[N]);
}
  
// Driver Code
public static void main(String[] args)
{
    int N = 21;
      
    findNthNumber(N);
}
}
  
// This code is contributed by Amit Katiyar

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Python3

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# Python 3 Program to find Nth number with
# absolute difference between all 
# adjacent digits at most 1.
  
# Return Nth number with
# absolute difference between all 
# adjacent digits at most 1.
def findNthNumber(N):
      
    # To store all such numbers
    arr = [0 for i in range(N + 1)]
      
    q = []
  
    # Enqueue all integers from 1 to 9 
    # in increasing order.
    for i in range(1, 10, 1):
        q.append(i)
  
    # Perform the operation N times so that
    # we can get all such N numbers.
    for i in range(1, N+1, 1):
          
        # Store the front element of queue,
        # in array and pop it from queue.
        arr[i] = q[0]
        q.remove(q[0])
  
        # If the last digit of dequeued integer is
        # not 0, then enqueue the next such number.
        if (arr[i] % 10 != 0):
            q.append(arr[i] * 10 + arr[i] % 10 - 1)
  
        # Enqueue the next such number
        q.append(arr[i] * 10 + arr[i] % 10)
  
        # If the last digit of dequeued integer is
        # not 9, then enqueue the next such number.
        if (arr[i] % 10 != 9):
            q.append(arr[i] * 10 + arr[i] % 10 + 1)
      
    print(arr[N])
  
# Driver Code
if __name__ == '__main__':
      
    N = 21
    findNthNumber(N)
  
# This code is contributed by Samarth

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Output:

45

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Improved By : ipg2016107, amit143katiyar