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.
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.
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Improved By : Sam007