$C o n s i d e r t h e f o l l o w i n g f u n c t i o n s, f o o () a n d b a r () : T h e t i m e c o m p l e x i t y o f t h e f u n c t i o n s f o o () a n d b a r () i s i n t f o o (i n t n) {i f (n < = 1) r e t u r n 2^{n}; r e t u r n 8 * f o o (n - 1);} i n t b a r (i n t n) {i f (n < = 1) r e t u r n 3^{n}; r e t u r n 3^{} b a r (n - 1) + 2 b a r (n - 1);}$

O (2^{n}) f o r f o o () a n d O (3^{n}) f o r b a r ()

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O (n) f o r f o o () a n d O (2^{n}) f o r b a r ()

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O (2^{n}) f o r b o t h f o o () a n d b a r ()

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O (n) f o r b o t h f o o () a n d b a r ()

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Solution

The correct option is B $O (n) f o r f o o () a n d O (2^{n}) f o r b a r ()$
$T h e r e c u r r e n c e r e l a t i o n f o r f o o (n) : T (n) = T (n - 1) + c; w h e r e c i s a constant \Rightarrow T (n) = O (n) T h e t i m e r e c u r r e n c e f o r b a r (n); T (n) = 2 T (n - 1) + c; c i s a constant \Rightarrow T (n) = O (2^{n})$

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Similar questions

int fun(int n) {static int x =1; x = x - n; (x < 10) return (fun(fun(n - 1))); else return x;}

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\sqrt{n}

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