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++

// 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;
}

C

// C program to find
// the N-th term in the series
// 1, 2, 11, 12, 21...
#include <stdio.h>

// 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
    printf("%d",printNthElement(N));

    return 0;
}

// This code is contributed by kothavvsaakash.

Java

// 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));
    }
}

Python3

# 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))

C#

// 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

PHP

<?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
?>

Javascript

<script>
    // Javascript 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)
        let arr = new Array(n + 1);
        arr[1] = 1;
        arr[2] = 2;

        for (let i = 3; i <= n; i++) {
            // If i is odd
            if (i % 2 != 0)
                arr[i] = arr[parseInt(i / 2, 10)] * 10 + 1;
            else
                arr[i] = arr[parseInt(i / 2, 10) - 1] * 10 + 2;
        }
        return arr[n];
    }

    let n = 5;

      document.write(printNthElement(n));

// This code is contributed by vaibhavrabadiya117.
</script>

Output:

21

Time Complexity: O(N)

Auxiliary Space: O(N)