Let n be any power raised to base 2 i.e 2n. We are given the number n and our task is to find out the number of digits contained in the number 2n.
Input : n = 5 Output : 2 Explanation : 2n = 32, which has only 2 digits. Input : n = 10 Output : 4 Explanation : 2n = 1024, which has only 4 digits.
We can write 2n using logarithms as:
2n = 10nlog102
Now suppose, x = nlog102,
Therefore, 2n = 10x
Also, we all know that the number, 10n will have (n+1) digits. Therefore, 10x will have (x+1) digits.
Or, we can say that 2n will have (x+1) digits as 2n = 10x.
Therefore, number of digits in 2n = (nlog102) + 1
Below is the implementation of above idea:
Time Complexity: O(n)
Auxiliary Space: O(1)
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