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1
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1
Fibonacci Series Complexity
PLEASE EXPLAIN HOW TO APPROACH THESE KIND OF PROBLEMS
PLEASE EXPLAIN HOW TO APPROACH THESE KIND OF PROBLEMS
669
views
answered
Nov 6, 2017
Algorithms
time-complexity
algorithms
asymptotic-notation
fibonacci-sequence
test-series
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–
1
votes
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 ?
$T(n)=4 T(n/2) + n^2 \sqrt{2}$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 +...
1.4k
views
answered
Nov 6, 2017
Algorithms
time-complexity
algorithms
recurrence-relation
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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 $n$ $n^2$ $n \log n$ $n \log^2n$
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$
33.2k
views
answered
Nov 6, 2017
Algorithms
gatecse-2004
algorithms
sorting
asymptotic-notation
easy
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