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#time complexity
amit166
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Nov 24, 2018
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Lakshman Bhaiya
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T(n)=4T(n/2)+n$^{2}\sqrt{2}$ T.C
algorithms
recurrence-relation
amit166
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Nov 24, 2018
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Lakshman Bhaiya
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Shubhanshu
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Nov 24, 2018
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$f(n)$ are given in asymptotic notation, so you can drop that constant.
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goxul
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Nov 24, 2018
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Substitution is also easy. After $k$ iterations, the recurrence will be $T(n) = 4^kT(n/2^k) + \sqrt 2 k n^2$, where $k = \log n$, making it $O(n^2 \log n)$.
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T(n)=mT(n/2)+an$^{2}$ what time complexity
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Comparing with masters theorem
T(n)=aT(n/b)+f(n)
a=4 , b=2 n^(log a base b) is n^2
f(n) = n^2 * root(2) which is asymptotically equal to n^2 so by case 3 of masters theorem
Theta(n log a base b *(log n)^ k+1)
K is the power in f(n) ... in ques it is 0
Sachdev aprajita
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May 22, 2019
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