Stack and Queue

Stack From Two Queues Describe how to implement the stack ADT using two queues The solution should only use the standard queue operations: ENQUEUE, DEQUEUE, IS-EMPTY What will be the running times of the PUSH and POP operations?

Stack From Two Queues – Solution We will call the two queues Q1 and Q2 Additionally, we will hold a variable called curQueue, which points to the queue that holds the last value pushed to our stack Initialization: curQueue  Q1

Stack From Two Queues – Solution (continued) IS-EMPTY return IS-EMPTY (curQueue) PUSH (val) ENQUEUE (curQueue, val) TOP val  POP return val

Stack From Two Queues – Solution (continued) POP otherQueue  (curQueue = Q1)? Q2: Q1 while (!IS-EMPTY (curQueue)) temp  DEQUEUE (curQueue) if (!IS-EMPTY (curQueue)) // Not last? ENQUEUE (otherQueue, temp) curQueue  otherQueue return temp

Stack From Two Queues – Solution Complexity Since PUSH is simply implemented as a single ENQUEUE operation, its running time is O(1) For POP operations, we always need to transfer the complete stack contents from one queue to the other, therefore performing a POP takes O(n) time

Two Stacks in a Single Array Describe how to implement two stacks inside a single array The total number of elements in both stacks is limited by the array length Therefore, just splitting the array into two equally-sized parts will not do All stack operations should run in O(1)

Two Stacks in a Single Array – Solution Concept We are given an array A with n elements, numbered 0 to n – 1 We define two markers t1 and t2 that point to the next free cell in stacks S1 and S2 respectively Initialization: t1 = 0 t2 = n – 1

Two Stacks in a Single Array – Solution Concept S1 grows up from the beginning of the array, S2 grows down from its end We know that the array is full when the two markers switch places (t1 > t2) The complexity is O(1) for all operations The array usage is optimal

Two Stacks in a Single Array – Solution Visualization 7 3 2 9 5 6 t1 t2

Two Stacks in a Single Array – Solution Details S1.IS-EMPTY if (t1 = 0) return true else return false S2.IS-EMPTY if (t2 = n - 1)

Two Stacks in a Single Array – Solution Details (continued) S1.PUSH (val) if (t1 > t2) error "stack overflow" A[t1] = val t1++ S2.PUSH (val) A[t2] = val t2--

Two Stacks in a Single Array – Solution Details (continued) S1.POP if (t1 = 0) error "stack underflow" t1-- return A[t1] S2.POP if (t2 = n - 1) t2++ return A[t2]

Two Stacks in a Single Array – Solution Details (continued) S1.TOP if (t1 = 0) error "stack is empty" return A[t1] S2.TOP if (t2 = n - 1) return A[t2]

Evaluating Arithmetic Expressions Fully parenthesized arithmetic expressions can be evaluated using Dijkstra's two-stack algorithm Uses two separate stacks – one for the operators and another for the operands See Java code in: Evaluate.java.html

Deque (or Double-Ended Queue) A deque is a generalization of the Queue ADT, that supports reading and writing from both ends of the queue We would like to support an input-restricted deque : Deletion can be made from both ends Input can only be made at one end

Deque We note that an input-restricted deque is a combination of a queue and a stack We use a standard queue, and modify it to also support PUSH and POP The tail (or rear) of the queue will also be used as the stack top

Deque Implementing PUSH is now trivial – simply call ENQUEUE POP is very similar to DEQUEUE, accept that it returns tail instead of head The same IS-EMPTY function will work for both the stack and the queue Check if head = tail

Deque ENQUEUE PUSH POP DEQUEUE 9 5 6 2 3 head (front) tail (rear)