Find position of the given number among the numbers made of 4 and 7

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 constructed by 4, 7 digit only, we need to find position of this number in this series.

Examples:

Input : 7
Output : pos = 2 

Input : 444
Output : pos = 7



It is reverse of the following article :
Find n-th element in a series with only 2 digits (4 and 7) allowed | Set 2 (log(n) method)

                      ""
               /              \
             1(4)            2(7)
          /        \       /      \ 
        3(44)    4(47)   5(74)    6(77)
       / \       / \      / \      / \

The idea is based on the fact that all even positioned numbers have 7 as last digit and all odd positioned numbers have 4 as last digit.

If number is 4 then it is the left node of the tree then it corresponds to (pos+2)+1. Else right child node(7) corresponds to (pos*2)+2.

C++

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// C++ program to find position of a number
// in a series of numbers with 4 and 7 as the
// only digits.
#include <iostream>
#include <algorithm>
using namespace std;
  
int findpos(string n)
{
    int i = 0, pos = 0;
    while (n[i] != '\0') {
  
        // check all digit position
        switch (n[i]) 
        {
  
        // if number is left then pos*2+1
        case '4':
            pos = pos * 2 + 1;
            break;
  
        // if number is right then pos*2+2
        case '7':
            pos = pos * 2 + 2; 
            break;
        }
        i++;
    }
    return pos;
}
  
// Driver code
int main()
    // given a number which is constructed 
    // by 4 and 7 digit only 
    string n = "774"
    cout << findpos(n);
    return 0;
}

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Java

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// java program to find position of a number
// in a series of numbers with 4 and 7 as the
// only digits.
import java.util.*;
  
class GFG {
      
    static int findpos(String n)
    {
          
        int k = 0, pos = 0, i = 0;
        while (k != n.length()) {
  
            // check all digit position
            switch (n.charAt(i)) {
  
            // if number is left then pos*2+1
            case '4':
                pos = pos * 2 + 1;
                break;
  
            // if number is right then pos*2+2
            case '7':
                pos = pos * 2 + 2;
                break;
            }
              
            i++;
            k++;
        }
          
        return pos;
    }
  
    // Driver code
    public static void main(String[] args)
    {
          
        // given a number which is constructed
        // by 4 and 7 digit only
        String n = "774";
          
        System.out.println(findpos(n));
    }
}
  
// This code is contributed by Sam007.

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Python3

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# python program to find position 
# of a number in a series of 
# numbers with 4 and 7 as the
# only digits.
def findpos(n):
    i = 0
    j = len(n)
    pos = 0
    while (i<j):
          
        # check all digit position
        # if number is left then 
        # pos*2+1
        if(n[i] == '4'):
            pos = pos * 2 + 1
              
        # if number is right then
        # pos*2+2
        if(n[i] == '7'):
            pos = pos * 2 + 2
          
        i= i+1
      
    return pos
  
  
# Driver code
# given a number which is constructed 
# by 4 and 7 digit only 
n = "774"
print(findpos(n))
  
# This code is contributed by Sam007

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C#

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// C# program to find position of 
// a number in a series of numbers
// with 4 and 7 as the only digits.
using System;
  
class GFG
{
    static int findpos(String n)
    {
          
        int k = 0, pos = 0, i = 0;
        while (k != n.Length) {
  
            // check all digit position
            switch (n[i]) {
  
            // if number is left then pos*2+1
            case '4':
                pos = pos * 2 + 1;
                break;
  
            // if number is right then pos*2+2
            case '7':
                pos = pos * 2 + 2;
                break;
            }
              
            i++;
            k++;
        }
          
        return pos;
    }
  
    // Driver code
    static void Main()
    {
          
        // given a number which is constructed
        // by 4 and 7 digit only
        String n = "774";
          
        Console.Write(findpos(n));
    }
      
}
  
// This code is contributed by Sam007

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PHP

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<?php
// PHP program to find position of a number
// in a series of numbers with 4 and 7 as the
// only digits.
  
function findpos($n)
{
    $i = 0;
    $pos = 0;
    while($i < strlen($n)) {
  
        // check all digit position
        switch ($n[$i]) 
        {
  
        // if number is left then pos*2+1
        case '4':
            $pos = $pos * 2 + 1;
            break;
  
        // if number is right then pos*2+2
        case '7':
            $pos = $pos * 2 + 2; 
            break;
        }
        $i++;
    }
    return $pos;
}
  
    // Driver code
    // given a number which
    // is constructed by 4 
    // and 7 digit only 
    $n = "774"
    echo findpos($n);
      
// This code is contributed by Sam007
?>

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Output:

13

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

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Improved By : Sam007