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Interview Question for Software Engineer at
Google:
Jul 31, 2012
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2 of 3 people found this helpful
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Star with a = 1 and check if a*a is > N. If so you return a-1 else a++ and move forward.
The fact that here we cannot use the floating point calculation makes everything trivial and not much accurate.
Anyway convert this strategy into an accurate one using floating point it's easy as well.
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1 of 1 people found this helpful
X(n+1) = (X(n)^2 + N) / (2 * X(n))
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3 of 3 people found this helpful
http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method
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