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# 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.
This article is contributed by Devanshu Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.