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. 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.

Implementation:

13

Time Complexity: O(n), where n represents the size of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.