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How to determine the time complexity of this loop?
// func() is any constant root function for (int i = n; i > 0; i = func(i)) { // some O(1) expressions or statements } "In this case, i takes values n, n1/k, (n1/k)1/k = n1/k2, ... do we calculate that there are logk(log(n)) iterations? Source: http://www.geeksforgeeks.org/time-complexity-loop-loop-variable-expands-shrinks-exponentially/
// func() is any constant root functionfor (int i = n; i 0; i = func(i)){ // some O(1) expressions or statements}"In this case, i takes values n, n1/k, (n1/k)1/k = n1/...
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Nov 7, 2017
Algorithms
algorithms
asymptotic-notation
time-complexity
space-complexity
non-gate
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1
answer
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 +...
1.5k
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commented
Nov 6, 2017
Algorithms
time-complexity
algorithms
recurrence-relation
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2
answers
3
Fibonacci Series Complexity
PLEASE EXPLAIN HOW TO APPROACH THESE KIND OF PROBLEMS
PLEASE EXPLAIN HOW TO APPROACH THESE KIND OF PROBLEMS
671
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answer edited
Nov 6, 2017
Algorithms
time-complexity
algorithms
asymptotic-notation
fibonacci-sequence
test-series
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–
5
answers
4
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$
33.3k
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answered
Nov 6, 2017
Algorithms
gatecse-2004
algorithms
sorting
asymptotic-notation
easy
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