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What happens if we apply the below operations on an input sequence? i. construct a cartesian tree for input sequence ii. put the root element of above tree in a priority queue iii. if( priority queue is not empty) then iv. search and delete minimum value in priority queue v. add that to output vi. add cartesian tree children of above node to priority queue
A double-ended queue supports operations like adding and removing items from both the sides of the queue. They support four operations like addFront(adding item to top of the queue), addRear(adding item to the bottom of the queue), removeFront(removing item from the top of the queue) and removeRear(removing item from the bottom of the queue). You are given only stacks to implement this data structure. You can implement only push and pop operations. What’s the time complexity of performing addFront and addRear? (Assume ‘m’ to be the size of the stack and ‘n’ to be the number of elements)
A Double-ended queue supports operations such as adding and removing items from both the sides of the queue. They support four operations like addFront(adding item to top of the queue), addRear(adding item to the bottom of the queue), removeFront(removing item from the top of the queue) and removeRear(removing item from the bottom of the queue). You are given only stacks to implement this data structure. You can implement only push and pop operations. What are the total number of stacks required for this operation?(you can reuse the stack)
A queue is implemented using a non-circular singly linked list. The queue has a head pointer and a tail pointer, as shown in the figure. Let n denote the number of nodes in the queue. Let enqueue be implemented by inserting a new node at the head, and dequeue be implemented by deletion of a node from the tail. Which one of the following is the time complexity of the most time-efficient implementation of enqueue and dequeue, respectively, for this data structure?
Which of the following statements are true? i) practical application of XOR linked lists are in environments with limited space requirements, such as embedded devices. ii)xor lists are not suitable because most garbage collectors will fail to work properly with classes or structures that don’t contain literal pointers iii)in order to calculate the address of the next node you need to remember the address of the previous node iv)xor lists are much efficient than single, doubly linked lists and arrays
Consider a standard circular queue ‘q’ implementation (which has same condition for queue full and queue empty) whose size is 11 and the elements of the queue are q[0], q[1], .... q[10]. The front and rear pointers are initialized to point at q[2]. In which position will the ninth element be added ?
If a higher priority process arrives and wants service, the memory manager can swap out the lower priority process to execute the higher priority process. When the higher priority process finishes, the lower priority process is swapped back in and continues execution. This variant of swapping is sometimes called?
In priority scheduling algorithm, when a process arrives at the ready queue, its priority is compared with the priority of ____________
A circular queue has been implemented using a singly linked list where each node consists of a value and a single pointer pointing to the next node. We maintain exactly two external pointers FRONT and REAR pointing to the front node and the near node of the queue, respectively. Which of the following statements is/are CORRECT for such a circular queue, so that insertion and deletion operations can be performed in O(1) time? I. Next pointer of front node points to the rear node. II. Next pointer of rear node points to the front node.
A scheduling algorithm assigns priority proportional to the waiting time of a process. Every process starts with priority zero (the lowest priority). The scheduler re-evaluates the process priorities everyT time units and decides the next process to schedule. Which one of the following is TRUE if theprocesses have no I/O operations and all arrive at time zero?