This document introduces stacks and their implementation using arrays. It defines a stack as a data structure that follows the LIFO (last in, first out) principle, where the last element added is the first removed. It describes how stacks can be modeled as a plate or coin stacker, with elements pushed onto the top and popped off the top. The key stack operations of push and pop are introduced, with push adding an element to the top and pop removing the top element. The implementation of stacks using arrays is then covered, with an array representing the stack and a variable tracking the top index. Pseudocode is provided for push and pop operations on the array-based stack.

1. Introduction to STACK
Ravi
presents ...
...

2. Stack
A stack is used to store elements where
the Last element In is the First one Out
(LIFO).
A common model of a stack is a plate or
coin stacker.
Plates are "pushed" onto to the top and
“pooped” off the top.

3. Stack
New elements are added or pushed onto
the top of the stack.
The first element to be removed or
popped is taken from the top - the last
one in.

4. Stack

5. Stack Operations
A stack is generally implemented with only two
principle operations
Push adds an item to a stack
Pop extracts the most recently pushed item from the
stack
Other methods such as
Top returns the item at the top without removing it
Isempty determines whether the stack has anything in it

6. Stack Implementation
Static Implementation (Using arrays)
Dynamic Implementation (Using dynamic
lists)

7. Stack Implementation
Using Arrays
For the static implementation of stack an
array will be used.
This array will hold the stack elements.
The top of a stack is represented by an
integer type variable which contains the
index of an array containing top element
of a stack.

8. Implementation of push operation on
stack
Void push(s,element)
1. If s Top=Max-1 print “Overflow stack” End If
 2. set s top=s top +1
 3. set s item[s top]=element
4. print “Element is pushed”
5. End

9. Stack Implementation
Using Arrays
4
3
2
1
0
Empty stack
StackSize = 5
top = -1
70
1
2
3
4
top
Push 7
70
81
2
3
4
top
Push 8 Push 9
70
81
92
3
4
top
Push 4
70
81
92
43
4
top
Push 5
70
81
92
43
54
top
top = StackSize – 1,
Stack is full,
We can’t push more elements.

10. Implementation of pop operation on
stack
int pop(s)
1. If s Top= -1 print “Underflow stack” return 0 and
goto step-5 End If
2. set popped=s item[s top]
3. set s top=s top -1
4. return popped
5. End

11. Stack Implementation Using Arrays
4
3
2
1
0
Empty stack
top = -1
We can’t pop mpre
elements
70
1
2
3
4
top
Pop
70
81
2
3
4
top
70
81
92
3
4
top
70
81
92
43
4
top
70
81
92
43
54
top
top = StackSize – 1,
Stack is full,
We can’t push more elements.
Pop
Pop
Pop Pop

12. THANK YOU