Skip to content
Related Articles

Related Articles

Improve Article
Find position of the given number among the numbers made of 4 and 7
  • Difficulty Level : Medium
  • Last Updated : 27 Apr, 2021

Consider a series of numbers composed of only digits 4 and 7. The first few numbers in the series are 4, 7, 44, 47, 74, 77, 444, .. etc. Given a number constructed by 4, 7 digits only, we need to find the 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 the last digit and all odd positioned numbers have 4 as the last digit.
If the 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++




// 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;
}

Java




// 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.

Python3




# 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

C#




// 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

PHP




<?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
?>

Javascript




<script>
    // Javascript program to find position of a number
// in a series of numbers with 4 and 7 as the
// only digits.
   
function findpos(n)
{
    let i = 0;
    let pos = 0;
    while(i < 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++;
    }
    return pos;
}
   
    // Driver code
    // given a number which
    // is constructed by 4
    // and 7 digit only
    let n = "774";
    document.write(findpos(n));
       
// This code is contributed by _saurabh_jaiswal
</script>

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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :