Solve recurence relation
for T(n) = 4T(n/2)+n
By master's method
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Answer:
T(n)=4T(n/2)+n2. For this recurrence, there are again a=4 subproblems, each dividing the input by b=2, but now the work done on each call is f(n)=n2.
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The master method
Case 1: f(n) is O(nlogba - ε). ...
Case 2: f(n) is Θ(nlogba). ...
Case 3: f(n) is Ω(nlogba + ε).
Explanation:
hope it's help you
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