in Algorithms
294 views
1 vote
1 vote

in Algorithms
294 views

3 Comments

$n^{2^{n}} > n! > 2^n$

$f_3 > f_1 > f_2$

so, option A is correct

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why we won't consider time complexity= $n^{2}*2^{n}$

if addition we will not consider but multiplication. i don't know exactly.

can u please explain
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$n!$ $n^2 \times 2^n$ $n^{2^n}$
  take log  
nlog n 2log n + nlog 2 2nlog n
nlog n 2log n + n 2nlog n
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