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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 ?>

Output:

774

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

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Improved By : Sam007, Mithun Kumar

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