The following statement is valid. log(n!) = 
(n log n).
(A)
True
(B)
False
Answer: (A)
Explanation:
Order of growth of \\log n! and n\\log n is the same for large values of , i.e., \\theta (\\log n!) = \\theta (n\\log n) . So time complexity of fun() is \\theta (n\\log n) . The expression \\theta (\\log n!) = \\theta (n\\log n) can be easily derived from following Stirling\’s approximation (or Stirling\’s formula). \\log n! = n\\log n – n +O(\\log(n))\\
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