\ Convert the following infix expression into postfix one: (A+B^D)/(D-F)+G PS: you have to illustrate the different steps of this conversion one by one.
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Q1 \ Convert the following infix expression into postfix one:
(A+B^D)/(D-F)+G
PS: you have to illustrate the different steps of this conversion one by one.
Step by step
Solved in 2 steps with 1 images
- There is one stacks and one queue q and then following operations are performed upon these two. Push A, enqueue B, push C, pop, pop, enqueue D, push E, dequeue, enqueue F .What is the total number of elements, which are left in the queue and stack in the end.501324-3 Fall 2021 Assignment Topics: Stack – Queue - Tree _____________________________________________________________ Question 1: Convert the following infix expression into postfix one: (A+B^D)/(D-F)+G PS: you have to illustrate the different steps of this conversion one by one. Token # Scan Postfix Expression Stack 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 The final solution is: …………………………………………………….. Question 2: Evaluate the postfix expression A B C D + * / Where A = 66, B = 1, C = 12 and D=10 Question 3: (Final Exam-Fall2018) Consider that you have a stack S and a queue Q. Draw S and Q after executing the following operations (Op). You have to indicate for each operation, the stack or the queue content in addition to the index value of the top (stack) or the front and the rear (queue)…Please answer the following, #3. Write clearly please. If postfix math is easy to analyze with a stack, would prefix math like ”/ * + 2 2 8 4” (which equals 8) be easy to analyze with a queue? How would you attempt to analyze this math with a queue? Lay out assumptions and use those in your solution as well.
- Implementing a Stack ADT using a linked solution. The implementations are linked Nodes. class Node: def __init__(self, data, node=None): # Initialize this node, insert data, and set the next node if any self.data = data self.chain = node class SubSetStack: def __init__(self, data=None): # Initialize this stack, and store data if it exists def push(self, data): # Add data to the beginning of the stack def pop(self): # Remove the element at the beginning of the stack. # Return the data in the element at the beginning of the stack, or None if the stack is empty def top(self): # Return the data in the element at the beginning but does not remove it. # Return None if stack is empty. def __len__(self): # Return the number of elements in the stack def subset_sum(target, sub_list): # Returns True if Target can be formed from sub_list repeated # some arbitrary number of times. In the above function a Node…Answer it fast and correct. Don't stuck. i ll rate Fill in the underlines below to complete a function template, reverseStack, that takes as a parameter a stack object and uses a queue object to reverse the elements of the stack. C++ code Answer: template <class Type> void reverseStack(stackType<Type> &s) { linkedQueueType<Type> q; Type elem; while (_________________) // The stack is not empty. { // Copy the top element of the stack into the queue //and remove it from the stack. elem = _________________; _________________; _________________; } while (_________________) // The queue is not empty. { // Copy the front element of the queue into the stack //and remove it from the queue. elem = _________________; _________________; _________________; } }10.7 Suppose you wish to reverse the order of elements of a queue. Usingonly Queue operations, describe how this would be done. (Hint: While you can’tuse a stack, you can use something similar.)10.8 Over time, the elements 1, 2, and 3 are pushed onto the stack, amongothers, in that order. What sequence(s) of popping the elements 1, 2 and 3 offthe stack are not possible?
- TRUE or FALSE? Please answer the following question and state the reason why: Thank you! In a Dynamic Stack, the pointer top stays at the head after a push operation. During a Pop operation in Static Stack, the elements are being moved one step up. In a dynamic implementation of stack, the pointer top has an initial value of null. STL empty function will yield a value of true if the stack has elements.A SpecialStackTM is a stack modified to support the following two operations: PUSHCLEAR(v) successively pops consecutive items from the top of the stack that are less than u, then pushes v onto the stack. POP() deletes the item at the top of the stack. Assume these operations are implemented using a singly-linked list. PUSHCLEAR(v) iter- ates through the linked list to pop the applicable items and then adds v to the beginning. POP() deletes the first item. (a) Describe the ordering of values on the stack. (b) Explain how a single call to PUSHCLEAR(V) could take (n) time. (c) If we assume that every operation takes linear time, we get a naive bound of O(n²) on the total runtime. But this doesn't consider the fact that these expensive operations happen infrequently so let's analyze it more closely. Show that any sequence of n operations takes O(n) time. -Short answer Answer the following in your own words: Your friend says they implemented a stack as a linked list with reference pointers to both a head and the tail node (i.e. they implemented a stack as a double ended linked list). They said this significantly improved the performance for all functions (push, pop, search, and update). Are they correct? Why or why not? Another friend implemented a queue as a linked list with only a head pointer (no tail node). She said this made performance of all functions (insert, remove, search, and update) categorically equal to a queue as a linked list with a head and a tail node. Is she correct? Why or why not?
- 2. Given the following stack A = { 29,18,10,15,20,9,5,13,2,4,15} Create a queue by taking the elements from the top of the stack and adding them to a queue2. An ordered stack is a data structure that stores a sequence of items and supports the following operations. • OrderedPush(x): removes all items smaller than x from the beginning of the sequence and then adds x to the beginning of the sequence. • Pop: deletes and returns the first item in the sequence (or Null if the sequence is empty). Suppose we implement an ordered stack with a simple linked list, using the obvious OrderedPush and Pop algorithms. Prove that if we start with an empty data structure, the amortized cost of each OrderedPush or Pop operation is O(1).What would be the contents of the initially empty queue Q1 and stack S1 (show the results from left to right where the left side represents the bottom in case of the stack and the rear in case of the queue) after the following code is executed : S1=Stack.Stack() Q1=Queue.Queue() aList=[59, 8, 30, 4, 9, 15, 16, 2, 31, 14, 18] for i in aList: if i==16 or i==9: S1.pop() Q1.dequeue() Q1.enqueue(i) elif i>14: S1.push(i) else: Q1.enqueue(i)