Arrays/Stacks/Queues

Single correct

• ABoth operations can be performed in $O(1)$ time
• BAt most one operation can be performed in $O(1)$ time but the worst case time for the other operation will be $\Omega(n)$
• CThe worst case time complexity for both operations will be $\Omega(n)$
• DWorst case time complexity for both operations will be $\Omega(\log n)$

Answer key coming soon

while Q is not Empty do
     if S is Empty OR Top(S) ≤ Head(Q) then
          x := Dequeue(Q);
          Push(S,x);
     else
          x := Pop(S);
          Enqueue(Q,x);
     end
end

Numerical answer

Answer key coming soon