EBK DATA STRUCTURES AND ALGORITHMS IN C
4th Edition
ISBN: 9781285415017
Author: DROZDEK
Publisher: YUZU
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Question
Chapter 4, Problem 3E
Program Plan Intro
Stack:
Stack is a container in which elements inserted and removed in LIFO (Last In First Out) manner.
- “top” is address of top element of stack.
- Basic stack operations are given below:
- push(): Insert an object into stack top.
- pop(): Delete object in stack top.
- top(): Get object in stack top.
Explanation of Solution
- b. By not using additional stack:
Given stacks are “S1” and “S2”. Additional non-array variables are “Num”, “Num2” and “last_element”.
Contents of “S1” can be transfer into “S2” as same order in “S1” as shown below:
- “Num” is number of elements initially in stack “S1”.
- Decrement “Num” by one.
- Till “Num” becomes zero
- Equate “Num” into “Num2”.
- Till “Num2” becomes zero
- Pop each element from “S1” and push it into “S2”...
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Chapter 4 Solutions
EBK DATA STRUCTURES AND ALGORITHMS IN C
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- Write a procedure to remove the nth element (from the top) of a linked stack with the rest of the elements unchanged. Contrast this with a sequential stack implementation for the same problemarrow_forwardPlease assume that you have been given an implementation of a stack that supports both push and pop in O(1) time. With this information, you would like to implement a queue with these stacks. (a) In what way can you efficiently implement a queue using two of these stacks? “Efficiently” in this case means in a way which will allow you to do part B. (b) Please prove that the amortized cost of each dequeue and enqueue operation is O(1) for your stack-based queue by using the aggregate amortized analysis technique.arrow_forwardFind out how effectively the stack operates when it is given the opportunity to work in line with its inherent inclinations.arrow_forward
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