Advance test algorithms

Question

 

Comments

Take any value of x and see how many times fun1 is called. y doesn't matter here.

If x=4 then fun1(4,y) --> fun1(3,4+y) -->fun1(2,3+y)--->fun1(1,2+y)--->fun1(0,1+y)

5 times.

Similarly for any value of x fun1 will be called for x+1 times and in each call only constant work is done like comparison. So O(c*(x+1))=O(x)

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whatever the size of X you take the  recursion stack size or depth of the recursion tree is also X

try it !

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Ok bhai thnxx :)

Answer 1

T(n)=T(n-1)+1

complexity will be 1+1+1+1.....n times

=O(n)

so B is the answer