 Solution

Use two permanent stacks, $F$ and $B$, and one temporary stack, $T$. At operation boundaries $T$ is empty. Maintain the representation invariant that the deque, read from front to back, is the top-to-bottom contents of $F$ followed by the bottom-to-top contents of $B$. Thus the tops of $F$ and $B$ are the two accessible ends. Maintain their sizes using two counters, requiring only $O(1)$ additional storage in Big O notation.

Adding at the front pushes onto $F$; adding at the back pushes onto $B$. Removing at an end whose permanent stack is nonempty simply pops that stack. If both permanent stacks are empty, report an empty deque.

Suppose a front removal finds $F$ empty and $B$ contains $m>0$ elements. Rebalance as follows:

The parked elements are the back half of the deque; the unparked elements are the front half. Moving the front half onto $F$ reverses their stack order and exposes the correct front element. Parking and restoring the back half reverses it twice, preserving its original order in $B$. The temporary stack is empty again. For a back removal with $B$ empty, exchange the roles of $F$ and $B$ throughout. Rebalance approximately half the elements, so both ends retain accessible elements after a large transfer.

For the amortized analysis, count each push or pop as one unit and use the potential method withInitially the deque is empty, so $\Phi =0$, and $\Phi$ is always nonnegative. An ordinary insertion or removal costs one unit and changes the size difference by at most one, giving amortized cost at most $4$.

For a rebalancing removal, let $h=\lfloor m/2\rfloor$. The parked half makes two transfers per element and the other half makes one; each transfer is a pop and a push. Including the final removal, the actual cost isBefore rebalancing the absolute size difference is $m$. Immediately after the split it is at most $1$, and after the removal it is at most $2$. HenceSize tests and counter updates change only these constants. Summing the amortized costs and using nonnegative final potential proves $O(1)$ amortized running time for all four operations, with three stacks and $O(1)$ additional storage. An individual rebalancing operation can still take linear time in the current size.