Consider the following recurrence:
Which one of the following is true?
(A) T(n) = (loglogn)
(B) T(n) = (logn)
(C) T(n) = (sqrt(n))
(D) T(n) = (n)
Explanation: This question can be solved by first change of variable and then Master Method.
Let n = 2^m T(2^m) = T(2^(m/2)) + 1 Let T(2^m) = S(m) S(m) = 2S(m/2) + 1
Above expression is a binary tree traversal recursion whose time complexity is (m). You can also prove using Master theorem.
S(m) = (m) = (logn) /* Since n = 2^m */
Now, let us go back to the original recursive function T(n)
T(n) = T(2^m) = S(m) = (Logn)
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