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Answers by Narasimhan

1 vote

1

Fibonacci Series Complexity
PLEASE EXPLAIN HOW TO APPROACH THESE KIND OF PROBLEMS

answered in Algorithms Nov 6, 2017

548 views

1 vote

2

Time Complexity
$T(n)=4 T(n/2) + n^2 \sqrt{2}$ I have solved this by back substitution .. and it forms equations of the form $4k T(n/2k) + k n^2 \sqrt 2$ its giving time complexity as n2 + n2 log2n the answer is Theta(n2.5). i have two questions .. a) how can we get Theta(n2.5). b) is n2 log2n Asymptotically faster than n2 ?

answered in Algorithms Nov 6, 2017

1.0k views

3 votes

3

GATE CSE 2004 | Question: 29
The tightest lower bound on the number of comparisons, in the worst case, for comparison-based sorting is of the order of $n$ $n^2$ $n \log n$ $n \log^2n$

answered in Algorithms Nov 6, 2017

28.3k views