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.

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Answer 1

Question

Chapter 4, Problem 1E

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.

Expert Solution

Explanation of Solution

a.

Given stack is “S” and the additional stacks are “S1” and “S2”.

The elements in stack “S” can be reverse as shown below:

First pop each element from “S” and push it into “S1” till “S” becomes empty.
Pop each element from “S1” and push it into “S2” till “S1” becomes empty.
Pop each element from “S2” and push it into “S” till “S2” becomes empty.
Now “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4.

First pop each element from “S” and push it into “S1”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 1

Then pop each element from “S1” and push it into “S2”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 2

Last pop each element from “S2” and push it into “S”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 3

Finally “S” contains in reverse order.

Queue:

Queue is another data structure in which insertion and removal of elements are in FIFO(First In First Out) manner.
Basic operations are given below:
- Enqueue: Insert element into back of queue.
- Dequeue: Remove item from queue’s front.

Explanation of Solution

b.

Given stack is “S” and additional queue is “Q”.

The elements in stack “S” can be reverse as shown below:

Pop each element from “S” and insert it into “Q” till “S” becomes empty.

Delete each element from “Q” and push it into “S” till “Q” becomes empty.

Now stack “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4.

First pop each element from “S” and insert it into “Q”.

Then delete each element from “Q” and push it into “S”.

Finally, “S” contains in reverse order.

Explanation of Solution

c.

Given stack is “S”, additional stack is “S1” and additional variables are “Num”, Num2 and “top_element”. The elements in stack “S” can be reverse as shown below.

“Num” is number of elements initially in stack “S”.

Decrement “Num” by one.

Till “Num” becomes zero.

Equate “Num” into “Num2”.

Pop stack top of “S” and store in “top_element”.

Pop “Num2” elements from “S” and push into “S1”.

Then push element in “top_element” into “S”.

Then pop each element from “S1” and push into “S” till “S1” becomes empty.

Decrement value of “Num” by one.

Finally “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4. Stack contains four elements. So, “Num” is 4. Decrement it by one.

First iteration:

“Num” is 3, pop stack top and store in “top_element”. Then pop other three elements from “S” and push into “S1”.

Then push element in “top_element” into “S” and pop each element from “S1” and push into “S”.

Second iteration:

“Num” is 2, pop stack top and store in “top_element”. Then pop other two elements from “S” and push into “S1”.

Push element in “top_element” into “S” and pop each element from “S1” and push into “S”.

Third iteration:

“Num” is 1, pop stack top and store in “top_element”. Then pop one element from “S” and push into “S1”.

Push element in “top_element” into “S” and pop each element from “S1” and push into “S”

Fourth iteration:

“Num” is 0, exit loop.

Finally, the stack “S” contains elements in reverse order.

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Answer 2

Question

Chapter 4, Problem 1E

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.

Expert Solution

Explanation of Solution

a.

Given stack is “S” and the additional stacks are “S1” and “S2”.

The elements in stack “S” can be reverse as shown below:

First pop each element from “S” and push it into “S1” till “S” becomes empty.
Pop each element from “S1” and push it into “S2” till “S1” becomes empty.
Pop each element from “S2” and push it into “S” till “S2” becomes empty.
Now “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4.

First pop each element from “S” and push it into “S1”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 1

Then pop each element from “S1” and push it into “S2”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 2

Last pop each element from “S2” and push it into “S”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 3

Finally “S” contains in reverse order.

Queue:

Queue is another data structure in which insertion and removal of elements are in FIFO(First In First Out) manner.
Basic operations are given below:
- Enqueue: Insert element into back of queue.
- Dequeue: Remove item from queue’s front.

Explanation of Solution

b.

Given stack is “S” and additional queue is “Q”.

The elements in stack “S” can be reverse as shown below:

Pop each element from “S” and insert it into “Q” till “S” becomes empty.

Delete each element from “Q” and push it into “S” till “Q” becomes empty.

Now stack “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4.

First pop each element from “S” and insert it into “Q”.

Then delete each element from “Q” and push it into “S”.

Finally, “S” contains in reverse order.

Explanation of Solution

c.

Given stack is “S”, additional stack is “S1” and additional variables are “Num”, Num2 and “top_element”. The elements in stack “S” can be reverse as shown below.

“Num” is number of elements initially in stack “S”.

Decrement “Num” by one.

Till “Num” becomes zero.

Equate “Num” into “Num2”.

Pop stack top of “S” and store in “top_element”.

Pop “Num2” elements from “S” and push into “S1”.

Then push element in “top_element” into “S”.

Then pop each element from “S1” and push into “S” till “S1” becomes empty.

Decrement value of “Num” by one.

Finally “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4. Stack contains four elements. So, “Num” is 4. Decrement it by one.

First iteration:

“Num” is 3, pop stack top and store in “top_element”. Then pop other three elements from “S” and push into “S1”.

Then push element in “top_element” into “S” and pop each element from “S1” and push into “S”.

Second iteration:

“Num” is 2, pop stack top and store in “top_element”. Then pop other two elements from “S” and push into “S1”.

Push element in “top_element” into “S” and pop each element from “S1” and push into “S”.

Third iteration:

“Num” is 1, pop stack top and store in “top_element”. Then pop one element from “S” and push into “S1”.

Push element in “top_element” into “S” and pop each element from “S1” and push into “S”

Fourth iteration:

“Num” is 0, exit loop.

Finally, the stack “S” contains elements in reverse order.

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Answer 3

Question

Chapter 4, Problem 1E

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.

Expert Solution

Explanation of Solution

a.

Given stack is “S” and the additional stacks are “S1” and “S2”.

The elements in stack “S” can be reverse as shown below:

First pop each element from “S” and push it into “S1” till “S” becomes empty.
Pop each element from “S1” and push it into “S2” till “S1” becomes empty.
Pop each element from “S2” and push it into “S” till “S2” becomes empty.
Now “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4.

First pop each element from “S” and push it into “S1”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 1

Then pop each element from “S1” and push it into “S2”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 2

Last pop each element from “S2” and push it into “S”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 3

Finally “S” contains in reverse order.

Queue:

Queue is another data structure in which insertion and removal of elements are in FIFO(First In First Out) manner.
Basic operations are given below:
- Enqueue: Insert element into back of queue.
- Dequeue: Remove item from queue’s front.

Explanation of Solution

b.

Given stack is “S” and additional queue is “Q”.

The elements in stack “S” can be reverse as shown below:

Pop each element from “S” and insert it into “Q” till “S” becomes empty.

Delete each element from “Q” and push it into “S” till “Q” becomes empty.

Now stack “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4.

First pop each element from “S” and insert it into “Q”.

Then delete each element from “Q” and push it into “S”.

Finally, “S” contains in reverse order.

Explanation of Solution

c.

Given stack is “S”, additional stack is “S1” and additional variables are “Num”, Num2 and “top_element”. The elements in stack “S” can be reverse as shown below.

“Num” is number of elements initially in stack “S”.

Decrement “Num” by one.

Till “Num” becomes zero.

Equate “Num” into “Num2”.

Pop stack top of “S” and store in “top_element”.

Pop “Num2” elements from “S” and push into “S1”.

Then push element in “top_element” into “S”.

Then pop each element from “S1” and push into “S” till “S1” becomes empty.

Decrement value of “Num” by one.

Finally “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4. Stack contains four elements. So, “Num” is 4. Decrement it by one.

First iteration:

“Num” is 3, pop stack top and store in “top_element”. Then pop other three elements from “S” and push into “S1”.

Then push element in “top_element” into “S” and pop each element from “S1” and push into “S”.

Second iteration:

“Num” is 2, pop stack top and store in “top_element”. Then pop other two elements from “S” and push into “S1”.

Push element in “top_element” into “S” and pop each element from “S1” and push into “S”.

Third iteration:

“Num” is 1, pop stack top and store in “top_element”. Then pop one element from “S” and push into “S1”.

Push element in “top_element” into “S” and pop each element from “S1” and push into “S”

Fourth iteration:

“Num” is 0, exit loop.

Finally, the stack “S” contains elements in reverse order.

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Answer 4

Question

Chapter 4, Problem 1E

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.

Expert Solution

Explanation of Solution

a.

Given stack is “S” and the additional stacks are “S1” and “S2”.

The elements in stack “S” can be reverse as shown below:

First pop each element from “S” and push it into “S1” till “S” becomes empty.
Pop each element from “S1” and push it into “S2” till “S1” becomes empty.
Pop each element from “S2” and push it into “S” till “S2” becomes empty.
Now “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4.

First pop each element from “S” and push it into “S1”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 1

Then pop each element from “S1” and push it into “S2”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 2

Last pop each element from “S2” and push it into “S”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 3

Finally “S” contains in reverse order.

Queue:

Queue is another data structure in which insertion and removal of elements are in FIFO(First In First Out) manner.
Basic operations are given below:
- Enqueue: Insert element into back of queue.
- Dequeue: Remove item from queue’s front.

Explanation of Solution

b.

Given stack is “S” and additional queue is “Q”.

The elements in stack “S” can be reverse as shown below:

Pop each element from “S” and insert it into “Q” till “S” becomes empty.

Delete each element from “Q” and push it into “S” till “Q” becomes empty.

Now stack “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4.

First pop each element from “S” and insert it into “Q”.

Then delete each element from “Q” and push it into “S”.

Finally, “S” contains in reverse order.

Explanation of Solution

c.

Given stack is “S”, additional stack is “S1” and additional variables are “Num”, Num2 and “top_element”. The elements in stack “S” can be reverse as shown below.

“Num” is number of elements initially in stack “S”.

Decrement “Num” by one.

Till “Num” becomes zero.

Equate “Num” into “Num2”.

Pop stack top of “S” and store in “top_element”.

Pop “Num2” elements from “S” and push into “S1”.

Then push element in “top_element” into “S”.

Then pop each element from “S1” and push into “S” till “S1” becomes empty.

Decrement value of “Num” by one.

Finally “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4. Stack contains four elements. So, “Num” is 4. Decrement it by one.

First iteration:

“Num” is 3, pop stack top and store in “top_element”. Then pop other three elements from “S” and push into “S1”.

Then push element in “top_element” into “S” and pop each element from “S1” and push into “S”.

Second iteration:

“Num” is 2, pop stack top and store in “top_element”. Then pop other two elements from “S” and push into “S1”.

Push element in “top_element” into “S” and pop each element from “S1” and push into “S”.

Third iteration:

“Num” is 1, pop stack top and store in “top_element”. Then pop one element from “S” and push into “S1”.

Push element in “top_element” into “S” and pop each element from “S1” and push into “S”

Fourth iteration:

“Num” is 0, exit loop.

Finally, the stack “S” contains elements in reverse order.

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Answer 5

Chapter 4, Problem 1E

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.

Expert Solution

Explanation of Solution

a.

Given stack is “S” and the additional stacks are “S1” and “S2”.

The elements in stack “S” can be reverse as shown below:

First pop each element from “S” and push it into “S1” till “S” becomes empty.
Pop each element from “S1” and push it into “S2” till “S1” becomes empty.
Pop each element from “S2” and push it into “S” till “S2” becomes empty.
Now “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4.

First pop each element from “S” and push it into “S1”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 1

Then pop each element from “S1” and push it into “S2”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 2

Last pop each element from “S2” and push it into “S”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 3

Finally “S” contains in reverse order.

Queue:

Queue is another data structure in which insertion and removal of elements are in FIFO(First In First Out) manner.
Basic operations are given below:
- Enqueue: Insert element into back of queue.
- Dequeue: Remove item from queue’s front.

Explanation of Solution

b.

Given stack is “S” and additional queue is “Q”.

The elements in stack “S” can be reverse as shown below:

Pop each element from “S” and insert it into “Q” till “S” becomes empty.

Delete each element from “Q” and push it into “S” till “Q” becomes empty.

Now stack “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4.

First pop each element from “S” and insert it into “Q”.

Then delete each element from “Q” and push it into “S”.

Finally, “S” contains in reverse order.

Explanation of Solution

c.

Given stack is “S”, additional stack is “S1” and additional variables are “Num”, Num2 and “top_element”. The elements in stack “S” can be reverse as shown below.

“Num” is number of elements initially in stack “S”.

Decrement “Num” by one.

Till “Num” becomes zero.

Equate “Num” into “Num2”.

Pop stack top of “S” and store in “top_element”.

Pop “Num2” elements from “S” and push into “S1”.

Then push element in “top_element” into “S”.

Then pop each element from “S1” and push into “S” till “S1” becomes empty.

Decrement value of “Num” by one.

Finally “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4. Stack contains four elements. So, “Num” is 4. Decrement it by one.

First iteration:

“Num” is 3, pop stack top and store in “top_element”. Then pop other three elements from “S” and push into “S1”.

Then push element in “top_element” into “S” and pop each element from “S1” and push into “S”.

Second iteration:

“Num” is 2, pop stack top and store in “top_element”. Then pop other two elements from “S” and push into “S1”.

Push element in “top_element” into “S” and pop each element from “S1” and push into “S”.

Third iteration:

“Num” is 1, pop stack top and store in “top_element”. Then pop one element from “S” and push into “S1”.

Push element in “top_element” into “S” and pop each element from “S1” and push into “S”

Fourth iteration:

“Num” is 0, exit loop.

Finally, the stack “S” contains elements in reverse order.

Answer 6

Chapter 4, Problem 1E

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.

Expert Solution

Explanation of Solution

a.

Given stack is “S” and the additional stacks are “S1” and “S2”.

The elements in stack “S” can be reverse as shown below:

First pop each element from “S” and push it into “S1” till “S” becomes empty.
Pop each element from “S1” and push it into “S2” till “S1” becomes empty.
Pop each element from “S2” and push it into “S” till “S2” becomes empty.
Now “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4.

First pop each element from “S” and push it into “S1”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 1

Then pop each element from “S1” and push it into “S2”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 2

Last pop each element from “S2” and push it into “S”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 3

Finally “S” contains in reverse order.

Queue:

Queue is another data structure in which insertion and removal of elements are in FIFO(First In First Out) manner.
Basic operations are given below:
- Enqueue: Insert element into back of queue.
- Dequeue: Remove item from queue’s front.

Answer 7

Chapter 4, Problem 1E

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.

Answer 8

Expert Solution

Explanation of Solution

a.

Given stack is “S” and the additional stacks are “S1” and “S2”.

The elements in stack “S” can be reverse as shown below:

First pop each element from “S” and push it into “S1” till “S” becomes empty.
Pop each element from “S1” and push it into “S2” till “S1” becomes empty.
Pop each element from “S2” and push it into “S” till “S2” becomes empty.
Now “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4.

First pop each element from “S” and push it into “S1”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 1

Then pop each element from “S1” and push it into “S2”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 2

Last pop each element from “S2” and push it into “S”.

EBK DATA STRUCTURES AND ALGORITHMS IN C, Chapter 4, Problem 1E , additional homework tip 3

Finally “S” contains in reverse order.

Queue:

Queue is another data structure in which insertion and removal of elements are in FIFO(First In First Out) manner.
Basic operations are given below:
- Enqueue: Insert element into back of queue.
- Dequeue: Remove item from queue’s front.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

b.

Given stack is “S” and additional queue is “Q”.

The elements in stack “S” can be reverse as shown below:

Pop each element from “S” and insert it into “Q” till “S” becomes empty.

Delete each element from “Q” and push it into “S” till “Q” becomes empty.

Now stack “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4.

First pop each element from “S” and insert it into “Q”.

Then delete each element from “Q” and push it into “S”.

Finally, “S” contains in reverse order.

Answer 13

Explanation of Solution

b.

Given stack is “S” and additional queue is “Q”.

The elements in stack “S” can be reverse as shown below:

Pop each element from “S” and insert it into “Q” till “S” becomes empty.

Delete each element from “Q” and push it into “S” till “Q” becomes empty.

Now stack “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4.

First pop each element from “S” and insert it into “Q”.

Then delete each element from “Q” and push it into “S”.

Finally, “S” contains in reverse order.

Answer 14

Explanation of Solution

c.

Given stack is “S”, additional stack is “S1” and additional variables are “Num”, Num2 and “top_element”. The elements in stack “S” can be reverse as shown below.

“Num” is number of elements initially in stack “S”.

Decrement “Num” by one.

Till “Num” becomes zero.

Equate “Num” into “Num2”.

Pop stack top of “S” and store in “top_element”.

Pop “Num2” elements from “S” and push into “S1”.

Then push element in “top_element” into “S”.

Then pop each element from “S1” and push into “S” till “S1” becomes empty.

Decrement value of “Num” by one.

Finally “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4. Stack contains four elements. So, “Num” is 4. Decrement it by one.

First iteration:

“Num” is 3, pop stack top and store in “top_element”. Then pop other three elements from “S” and push into “S1”.

Then push element in “top_element” into “S” and pop each element from “S1” and push into “S”.

Second iteration:

“Num” is 2, pop stack top and store in “top_element”. Then pop other two elements from “S” and push into “S1”.

Push element in “top_element” into “S” and pop each element from “S1” and push into “S”.

Third iteration:

“Num” is 1, pop stack top and store in “top_element”. Then pop one element from “S” and push into “S1”.

Push element in “top_element” into “S” and pop each element from “S1” and push into “S”

Fourth iteration:

“Num” is 0, exit loop.

Finally, the stack “S” contains elements in reverse order.

Answer 15

Explanation of Solution

c.

Given stack is “S”, additional stack is “S1” and additional variables are “Num”, Num2 and “top_element”. The elements in stack “S” can be reverse as shown below.

“Num” is number of elements initially in stack “S”.

Decrement “Num” by one.

Till “Num” becomes zero.

Equate “Num” into “Num2”.

Pop stack top of “S” and store in “top_element”.

Pop “Num2” elements from “S” and push into “S1”.

Then push element in “top_element” into “S”.

Then pop each element from “S1” and push into “S” till “S1” becomes empty.

Decrement value of “Num” by one.

Finally “S” contains elements in reverse order.

Example,

Initially “S” contains 1, 2, 3 and 4. Stack contains four elements. So, “Num” is 4. Decrement it by one.

First iteration:

“Num” is 3, pop stack top and store in “top_element”. Then pop other three elements from “S” and push into “S1”.

Then push element in “top_element” into “S” and pop each element from “S1” and push into “S”.

Second iteration:

“Num” is 2, pop stack top and store in “top_element”. Then pop other two elements from “S” and push into “S1”.

Push element in “top_element” into “S” and pop each element from “S1” and push into “S”.

Third iteration:

“Num” is 1, pop stack top and store in “top_element”. Then pop one element from “S” and push into “S1”.

Push element in “top_element” into “S” and pop each element from “S1” and push into “S”

Fourth iteration:

“Num” is 0, exit loop.

Finally, the stack “S” contains elements in reverse order.

Answer 16

Ch. 4 - Prob. 1E Ch. 4 - Prob. 2E Ch. 4 - Prob. 3E Ch. 4 - Prob. 4E Ch. 4 - Prob. 5E Ch. 4 - Prob. 6E Ch. 4 - Prob. 7E Ch. 4 - Prob. 8E Ch. 4 - Prob. 9E Ch. 4 - Prob. 10E

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.

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

Explanation of Solution

Explanation of Solution

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