Time complexity of fun()?

The inner loop does n iterations, then n/2, then n/4, etc. So the total number of inner loop iterations is:

   n + n/2 + n/4 + n/8 + ... + 1  
<= n * (1 + 1/2 + 1/4 + 1/8 + ...) 
 = 2n

(See Geometric series), and therefore is O(n).

For an input integer n,

for i=n, j loop will run n times, then for i= n/2, j loop will run n/2 times, . . so on, . . till i=1, where j loop will run 1 time.

So,

T(n) = (n + n/2 + n/4 + n/8 + ...... + 1) 
    T(n) = n(1 + 1/2 + 1/4 + 1/8 + ..... + 1/n)                     .....(1)
let 1/n = 1/2^k

so, k = logn

now, using summation of geometric series in eq 1

T(n) = n((1-1/2^k) / 1-1/2)
T(n) = 2n(1-1/2^k)

using k = logn
T(n) = 2n(1-1/2^logn)

now for larger value of n, logn tends to infinite and 1/2^infinite will tend to 0. so, T(n) = 2n so, T(n) = O(n)

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