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.
Input : n = 2 Output : 7 Input : n = 3 Output : 44 Input : n = 5 Output : 74 Input : n = 6 Output : 77
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.
In this code the total complexity is O(log n). Because while loop run log (n) times.
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