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Questions by prateekdwv
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Solve the recurrence relation $B(n) = 3B(n/log_2(n))+\theta(n)$
What is the solution of following recurrence relation. $B(2) = 1$ $B(n) = 3B(n/\log_2(n))+\Theta(n)$​
What is the solution of following recurrence relation.$B(2) = 1$$B(n) = 3B(n/\log_2(n))+\Theta(n)$​
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Mar 24, 2016
Algorithms
algorithms
time-complexity
recurrence-relation
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