# 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` `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

 `= 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

 ``

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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