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

Consider a series of numbers composed of only digits 4 and 7. First few numbers in the series are 4, 7, 44, 47, 74, 77, 444, .. 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

We have discussed a O(n) solution in below post. 
Find n-th element in a series with only 2 digits (4 and 7) allowed
In this post, a O(log n) solution is discussed which is based on below pattern in numbers. The numbers can be seen

                 ""
              /      \
            4         7
          /   \     /   \ 
        44    47   74    77
       / \   / \   / \  / \

The idea is to fill the required number from end. We know can observe that the last digit is 4 if n is odd and last digit is 7 if n is even. After filling last digit, we move to parent node in tree. If n is odd, then parent node corresponds to (n-1/2. Else parent node corresponds to (n-2)/2. 

C++

// C++ program to find n-th number containing
// only 4 and 7.
#include<bits/stdc++.h>
using namespace std;

string findNthNo(int n)
{
    string res = "";
    while (n >= 1)
    {
        // If n is odd, append 4 and
        // move to parent
        if (n & 1)
        {
            res = res + "4";
            n = (n-1)/2;
        }

        // If n is even, append 7 and
        // move to parent
        else
        {
            res = res + "7";
            n = (n-2)/2;
        }
    }

   // Reverse res and return.
   reverse(res.begin(), res.end());
   return res;
}

// Driver code
int main()
{
    int n = 13;
    cout << findNthNo(n);
    return 0;
}

Java

// java program to find n-th number
// containing only 4 and 7.
public class GFG {

    static String findNthNo(int n)
    {
        String res = "";
        while (n >= 1)
        {

            // If n is odd, append
            // 4 and move to parent
            if ((n & 1) == 1)
            {
                res = res + "4";
                n = (n - 1) / 2;
            }

            // If n is even, append
            // 7 and move to parent
            else
            {
                res = res + "7";
                n = (n - 2) / 2;
            }
        }

        // Reverse res and return.
        StringBuilder sb =
            new StringBuilder(res);
        sb.reverse();
        return new String(sb);
    }

    // Driver code
    public static void main(String args[])
    {
        int n = 13;

        System.out.print( findNthNo(n) );
    }
}

// This code is contributed by Sam007

Python3

# Python3 program to find
# n-th number containing
# only 4 and 7.
def reverse(s):
    if len(s) == 0:
        return s
    else:
        return reverse(s[1:]) + s[0]

def findNthNo(n):
    res = "";
    while (n >= 1):

        # If n is odd, append
        # 4 and move to parent
        if (n & 1):
            res = res + "4";
            n = (int)((n - 1) / 2);

            # If n is even, append7
            # and move to parent
        else:
            res = res + "7";
            n = (int)((n - 2) / 2);

    # Reverse res
    # and return.
    return reverse(res);

# Driver code
n = 13;
print(findNthNo(n));

# This code is contributed
# by mits

C#

// C# program to find n-th number
// containing only 4 and 7.
using System;
class GFG {

static string findNthNo(int n)
{
    string res = "";
    while (n >= 1)
    {

        // If n is odd, append 4 and
        // move to parent
        if ((n & 1) == 1)
        {
            res = res + "4";
            n = (n - 1) / 2;
        }

        // If n is even, append 7 and
        // move to parent
        else
        {
            res = res + "7";
            n = (n - 2) / 2;
        }
    }

    // Reverse res and return.
    char[] arr = res.ToCharArray();
    Array.Reverse(arr);
    return new string(arr);

}

// Driver Code
public static void Main()
{
        int n = 13;
        Console.Write( findNthNo(n) );
}
}

// This code is contributed by Sam007

PHP

<?php
// PHP program to find
// n-th number containing
// only 4 and 7.

function findNthNo($n)
{
    $res = "";
    while ($n >= 1)
    {
        // If n is odd, append
        // 4 and move to parent
        if ($n & 1)
        {
            $res = $res . "4";
            $n = (int)(($n - 1) / 2);
        }

        // If n is even, append
        // 7 and move to parent
        else
        {
            $res = $res . "7";
            $n = (int)(($n - 2) / 2);
        }
    }

// Reverse res
// and return.
return strrev($res);
}

// Driver code
$n = 13;
echo findNthNo($n);

// This code is contributed
// by mits
?>

Javascript

<script>

// javascript program to find n-th number
// containing only 4 and 7.

function findNthNo(n)
{
res = "";
while (n >= 1)
{

// If n is odd, append
// 4 and move to parent
if ((n & 1) == 1)
{
res = res + "4";
n = (n - 1) / 2;
}

// If n is even, append
// 7 and move to parent
else
{
res = res + "7";
n = parseInt((n - 2) / 2);
}
}

// Reverse res and return.
return res.split("").reverse().join("");
}

// Driver code
var n = 13;

document.write( findNthNo(n) );

// This code is contributed by 29AjayKumar

</script>

Output:

774

Time Complexity: O(logN), where N represents the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

In this code the total complexity is O(log n). Because while loop run log (n) times.
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