Consider the following operation along with Enqueue and Dequeue operations on queues, where k is a global parameter.    

     m = k 
     while ((Q is not empty) and (m > 0))  
     { Dequeue(Q) 
       m = m – 1 

What is the worst case time complexity of a sequence of n queue operations on an initially full  queue having n elements.?  

(A) $\Theta(n) $ 

(B) $\Theta(n+k)$

(C) $\Theta(nk)$

(D) $\Theta(n^2)$

Here suppose we do one dequeue operation, then the loop will run for min(n,k) times. Now remaining 1 operation can be 1 enqueue operation which will take O(1) time so total complexity in this case will be O(min(n,k)).

Suppose we have k=1 and do (n-1) dequeue operations then it will take k*(n-1) time for multideqeue function and remaining one enqueue operation will take O(1) time . So in total we are getting O(kn) time in this case.

I am confused on how to handle 'k' parameter when we are calculating complexity.

NOTE :- n queue operations can be any combination of enqueue/dequeue/multi-dequeue(as defined) operation.

Any help would be appreciated.

  • $\begingroup$ @Dukeling You can assume it n. I will include it in the question. $\endgroup$
    – Zephyr
    Dec 24, 2017 at 12:29

1 Answer 1


It's $\Theta(n)$. Observe that if you are limited by $n$ operations, the total number of elements that ever enter the queue is less than or equal to $n$. Furthermore, every element is processed at most twice: once when it enters the queue, and once when it leaves the queue, therefore the total time is at most a constant times $n$.

  • $\begingroup$ Also why is $\Theta(n+k)$ not the correct answer? $\endgroup$
    – Zephyr
    Dec 24, 2017 at 5:37
  • $\begingroup$ If at the beginning I perform 1 multi-D operation with k=n then it will take O(n) time. Now 'n-2' enqueue operations will take O(n-2) time. Now again 1 multideqeue operation will take O(n-2) time. So total O(3n-4). $\endgroup$
    – Zephyr
    Dec 24, 2017 at 6:13
  • $\begingroup$ The $k$ parameter is irrelevant to the complexity. Also, $\Theta(3n-4) = \Theta(n)$, I suggest you review the Landau notation. $\endgroup$
    – quicksort
    Dec 24, 2017 at 8:42
  • $\begingroup$ I know that constants are irrelevant in landau notation. Wouldn't $\Theta(n+k)$ be more precise ? $\endgroup$
    – Zephyr
    Dec 24, 2017 at 8:45
  • $\begingroup$ No. $k$ is irrelevant. $\endgroup$
    – quicksort
    Dec 24, 2017 at 8:46

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