Program to find N-th term of series 1, 2, 11, 12, 21….

Given a number N, the task is to find the Nth term of the series:

1, 2, 11, 12, 21…

Examples:

Input : N = 2
Output : 2

Input : N = 5
Output : 21


Approach:
The idea is based on the fact that the value of the last digit alternates in the series. For example, if the last digit of ith number is 1, then the last digit of (i-1)th and (i+1)th numbers must be 2.

Therefore, Create an array of size (n+1) and push 1 and 2(These two are always first two elements of series) to it.

Therefore the ith term of the array is:
1) If i is odd,
arr[i] = arr[i/2]*10 + 1;
2) If i is even,
arr[i] = arr[(i/2)-1]*10 + 2;

At last return arr[n].

Below is the implementation of the above idea:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find
// the N-th term in the series
// 1, 2, 11, 12, 21...
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find N-th number in series
int printNthElement(int N)
{
    // create an array of size (N+1)
    int arr[N + 1];
    arr[1] = 1;
    arr[2] = 2;
  
    for (int i = 3; i <= N; i++) {
        // If i is odd
        if (i % 2 != 0) {
  
            arr[i] = arr[i / 2] * 10 + 1;
        }
        else {
  
            arr[i] = arr[(i / 2) - 1] * 10 + 2;
        }
    }
    return arr[N];
}
  
// Driver code
int main()
{
  
    // Get N
    int N = 5;
  
    // Get Nth term
    cout << printNthElement(N);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find
// the N-th term in the series
// 1, 2, 11, 12, 21...
  
class FindNth {
  
    // Function to find n-th number in series
    static int printNthElement(int n)
    {
        // create an array of size (n+1)
        int arr[] = new int[n + 1];
        arr[1] = 1;
        arr[2] = 2;
  
        for (int i = 3; i <= n; i++) {
            // If i is odd
            if (i % 2 != 0)
                arr[i] = arr[i / 2] * 10 + 1;
            else
                arr[i] = arr[(i / 2) - 1] * 10 + 2;
        }
        return arr[n];
    }
  
    // main function
    public static void main(String[] args)
    {
        int n = 5;
  
        System.out.println(printNthElement(n));
    }
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to find
# the N-th term in the series
# 1, 2, 11, 12, 21...
  
# Return n-th number in series
def printNthElement(n) :  
          
    # create an array of size (n + 1)  
    arr =[0] * (n + 1);  
    arr[1] = 1
    arr[2] = 2
      
    for i in range(3, n + 1) :  
        # If i is odd  
        if (i % 2 != 0) :  
            arr[i] = arr[i // 2] * 10 + 1
        else :  
            arr[i] = arr[(i // 2) - 1] * 10 + 2
          
    return arr[n]  
          
# Driver code  
n = 5
print(printNthElement(n))  

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find
// the N-th term in the series
// 1, 2, 11, 12, 21...
using System;
  
class GFG 
{
  
// Function to find n-th
// number in series
static int printNthElement(int n)
{
    // create an array of size (n+1)
    int []arr = new int[n + 1];
    arr[1] = 1;
    arr[2] = 2;
  
    for (int i = 3; i <= n; i++)
    {
        // If i is odd
        if (i % 2 != 0)
            arr[i] = arr[i / 2] * 10 + 1;
        else
            arr[i] = arr[(i / 2) - 1] * 10 + 2;
    }
    return arr[n];
}
  
// Driver Code
public static void Main()
{
    int n = 5;
  
    Console.WriteLine(printNthElement(n));
}
}
  
// This code is contributed
// by inder_verma

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to find 
// the N-th term in the series 
// 1, 2, 11, 12, 21... 
  
// Function to find N-th 
// number in series 
function printNthElement($N
    // create an array of size (N+1) 
    $arr = array($N + 1);
    $arr[1] = 1; 
    $arr[2] = 2; 
  
    for ( $i = 3; $i <= $N; $i++)
    
        // If i is odd 
        if ($i % 2 != 0)
        
  
            $arr[$i] = $arr[$i / 2] * 
                            10 + 1; 
        
        else 
        
  
            $arr[$i] = $arr[($i / 2) - 1] *     
                                  10 + 2; 
        
    
    return $arr[$N]; 
  
// Driver code 
$N = 5; 
  
// Get Nth term 
echo printNthElement($N); 
  
// This code is contributed
// by Mahadev99
?>

chevron_right


Output:

21


My Personal Notes arrow_drop_up

pawanasipugmailcom

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : Mahadev99, inderDuMCA