Find n-th element in a series with only 2 digits (4 and 7) allowed

Consider a series of numbers composed of only digits 4 and 7. First few numbers in the series are 4, 7, 44, 47, 74, 44744,.. etc. Given a number n, we need to find n-th number in the series.
Examples: 

Input : n = 2
Output : 7

Input : n = 3
Output : 44

Input  : n = 5
Output : 74

Input  : n = 6
Output : 77

The idea is based on the fact that the value of last digit alternates in series. For example, if last digit of i-th number is 4, then last digit of (i-1)-th and (i+1)-th numbers must be 7.
We create an array of size (n+1) and push 4 and 7 (These two are always first two elements of series) to it. For more elements we check 
1) If i is odd, 
      arr[i] = arr[i/2]*10 + 4; 
2) If it is even, 
      arr[i] = arr[(i/2)-1]*10 + 7; 
At last return arr[n].

C++

// C++ program to find n-th number in a series
// made of digits 4 and 7
#include <bits/stdc++.h>
using namespace std;

// Return n-th number in series made of 4 and 7
int printNthElement(int n)
{
    // create an array of size (n+1)
    int arr[n+1];
    arr[1] = 4;
    arr[2] = 7;

    for (int i=3; i<=n; i++)
    {
        // If i is odd
        if (i%2 != 0)
            arr[i] = arr[i/2]*10 + 4;
        else
            arr[i] = arr[(i/2)-1]*10 + 7;
    }
    return arr[n];
}

// Driver code
int main()
{
    int n = 6;
    cout << printNthElement(n);
    return 0;
}

Java

// Java program to find n-th number in a series
// made of digits 4 and 7

class FindNth
{
    // Return n-th number in series made of 4 and 7
    static int printNthElement(int n)
    {
        // create an array of size (n+1)
        int arr[] = new int[n+1];
        arr[1] = 4;
        arr[2] = 7;

        for (int i=3; i<=n; i++)
        {
            // If i is odd
            if (i%2 != 0)
                arr[i] = arr[i/2]*10 + 4;
            else
                arr[i] = arr[(i/2)-1]*10 + 7;
        }
        return arr[n];
    }

    // main function
    public static void main (String[] args)
    {
        int n = 6;
        System.out.println(printNthElement(n));
    }
}

Python3

# Python3 program to find n-th number
# in a series made of digits 4 and 7

# Return n-th number in series made
# of 4 and 7
def printNthElement(n) :

    # create an array of size (n + 1)
    arr =[0] * (n + 1);
    arr[1] = 4
    arr[2] = 7

    for i in range(3, n + 1) :
        # If i is odd
        if (i % 2 != 0) :
            arr[i] = arr[i // 2] * 10 + 4
        else :
            arr[i] = arr[(i // 2) - 1] * 10 + 7

    return arr[n]

# Driver code
n = 6
print(printNthElement(n))

# This code is contributed by Nikita Tiwari.

C#

// C# program to find n-th number in a series
// made of digits 4 and 7
using System;

class GFG
{
    // Return n-th number in series made of 4 and 7
    static int printNthElement(int n)
    {
        // create an array of size (n+1)
        int []arr = new int[n+1];
        arr[1] = 4;
        arr[2] = 7;

        for (int i = 3; i <= n; i++)
        {
            // If i is odd
            if (i % 2 != 0)
                arr[i] = arr[i / 2] * 10 + 4;
            else
                arr[i] = arr[(i / 2) - 1] * 10 + 7;
        }
        return arr[n];
    }

    // Driver code
    public static void Main ()
    {
        int n = 6;
        Console.Write(printNthElement(n));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find n-th
// number in a series
// made of digits 4 and 7

// Return n-th number in
// series made of 4 and 7
function printNthElement($n)
{

    // create an array
    // of size (n+1)
    $arr[1] = 4;
    $arr[2] = 7;

    for ($i = 3; $i <= $n; $i++)
    {

        // If i is odd
        if ($i % 2 != 0)
            $arr[$i] = $arr[$i / 2] *
                               10 + 4;
        else
            $arr[$i] = $arr[($i / 2) - 1] *
                                    10 + 7;
    }
    return $arr[$n];
}

// Driver code
$n = 6;
echo(printNthElement($n));

// This code is contributed by Ajit.
?>

Javascript

<script>
// javascript program to find n-th number in a series
// made of digits 4 and 7


    // Return n-th number in series made of 4 and 7
    function printNthElement(n) {
        // create an array of size (n+1)
        var arr = Array(n + 1).fill(0);
        arr[1] = 4;
        arr[2] = 7;

        for (var i = 3; i <= n; i++) {
            // If i is odd
            if (i % 2 != 0)
                arr[i] = arr[i / 2] * 10 + 4;
            else
                arr[i] = arr[(i / 2) - 1] * 10 + 7;
        }
        return arr[n];
    }

    // main function

        var n = 6;
        document.write(printNthElement(n));

// This code is contributed by Princi Singh
</script>

Output: 

77

Time Complexity: O(n) since using a for loop
Auxiliary Space: O(n) for creating an array of size N + 1.

Find n-th element in a series with only 2 digits (4 and 7) allowed | Set 2 (log(n) method)
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