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

• Difficulty Level : Hard
• Last Updated : 31 Mar, 2021

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.#includeusing 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 codeint 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 coden = 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 Codepublic 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

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