While loop will run for the maximum number of times when the Queue elements are sorted in descending order.

Let's suppose that initially, Queue elements are 16, 15, 14, 13...............2, 1

Now, 16 will be first pushed into the stack, So, now stack top is 16 and Head(Q) is 15, So 16 will be popped out of the stack ( since, "if S is Empty OR Top(S) ≤ Head (Q) "returns false, hence else part will be executed) and enqueued into the Queue.

So, after two iterations Queue elements will be -> 15, 14, 13, .........................2, 1, 16 and stack will be empty.

Similarly, each of the elements 15,14,13...........2 will be pushed into the stack and immediately popped out of the stack(when popped out of the stack then also enqueue into the queue).So after 30 iterations stack will be empty and Queue contains will be like==> 1, 16, 15, 14, .....................2.

Now 1 will be Dequeued and pushed into the stack.Once 1 is pushed into the stack, it will never be popped(or we can say never be enqueued into the Queue again) because in order to Pop 1, there should be an element into the Queue which is less than 1 and that element comes at the front of the queue, since there is no element currently present in the Queue which is less than 1, there is no way to pop 1.

So, after 31 iterations Queue is==> 16, 15, 14, ........................2 and stack contains 1.

Now, the problem boils down to Queue with 15 elements.

Using the similar logic we can say after another 29 iterations (Total =31+29 )Queue will be like==> 16, 15, 14, ............3 and stack contains 1,2 (stack top is 2) and then 2 will remain there in the stack forever.

Similar way if we go on then, after 31 + 29 + 27 + 25 + ......+1 iterations Queue will be empty.

This is in A.P series with d=2. Sum = (16 *(1+31))/2= 16*32/2= 256