Given the following procedure, analyze its worst case time complexity in big-O notation by only considering the comparison operations.

The algorithm takes as input a list of n integers in non decreasing order and produces the list of all values that occur more than once. (Recall that a list of integers is non decreasing if each integer in the list is at least as large as the previous integer in the list.)

Comment: The outer while loop checks each element in the list. For each element, when a_j occurs more than once is identified, then the inner while loop will skip all element in the list equal to a_j so that move to the next element not equal to a_j (works because the list is sorted).

procedure duplicates(a1, a2,, an: integers in non decreasing order)

k := 0 {this counts the duplicates}

j := 2

while j n

if a_j = a_j1 then

k := k + 1

c_k := a_j

while j n and a_j = c_k

j := j + 1

j := j + 1

{c1, c2,, ck is the desired list}