IT2201 Data Structures

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PREPARING FOR THE BIT
IT2201 Data Structures Algorithms
dsa_at_ict.cmb.ac.lk
Preparing for BIT 17/05/2001
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STACKS

• A stack is a data structure in which all the
  access is restricted to the most recently
  inserted items.

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STACK MODEL
Input to a stack is by push

Access is by top

Deletion is by pop

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STACK MODEL
Pop,Top
Push
B
A
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STACK PROPERTIES

• The last item added to the stack is placed on the
  top and is easily accessible.
• Thus the stack is appropriate if we expect to
  access on the top item, all other items are
  inaccessible.

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Important stack applications

• Compiler Design
• Mathematical Expression Evaluation
• Balanced Spell Checker
• Simple Calculator

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IMPLEMENTATION OF STACKS

• There are two basic ways to arrange for constant
  time operations.
• The first is to store the items contiguously in
  an array.
• And the second is to store items non-contiguously
  in a linked list

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ARRAY IMPLEMENTATION

• To push an item into an empty stack, we insert it
  at array location 0. ( since all java arrays
  start at 0)
• To push the next item into at location 0 over to
  location to make room for new item.

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• This is easily done by defining an auxiliary
  integer variable known as stack pointer,which
  stores the array index currently being used as
  the top of the stack.
• A stack can be implemented with an array and an
  integer.
• The integer TOS (top of stack) provides the array
  index of the top element of the stack, when TOS
  is 1 , the stack is empty

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Stacks specifies two data fields such as

• The array ( which is expanded as needed stores
  the items in the stack)
• Topofstack (TOS) ( gives the index of the current
  top of the stack, if stack is empty , this index
  is 1.

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ALGORITHMS FOR PUSHING POPPING

• PUSH
• If stack is not full then
• Add 1 to the stack pointer.
• Store item at stack pointer location.

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ALGORITHMS FOR PUSHING POPPING

• POP
• If stack is not empty then
• Read item at stack pointer location.
• Subtract 1 from the stack pointer.

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How to stack routines workempty
stackpush(a),push(b)pop
TOS (0)
a
Push (a)
Stack is empty TOS(-1)
TOS (1)
b
TOS (0)
a
a
pop
Push (b)
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Java implementation

• Zero parameter constructor for array based stack
• Public stackar( )
• / construct the stack
• thearray new objectdefault-capacity
• Tos -1

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Isempty returns true if stack is empty, false
otherwise

• Isempty( )
• Public Boolean Isempty( )
• Return tos -1

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Isfull( ) returns true if stack is full , false
otherwise

• Isfull( )
• Public Boolean isfull( )
• Return tos stacksize-1 (default capacity)

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push method for array based stack

• Insert a new item into the stack
• Public void push (object x)
• If isfull( )
• Throw new stackexception ( stack is full)
• Thearraytosx

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Pop method for array based stack

• Remove the most recently inserted item from the
  stack
• Exception underflow if the stack is empty
• Public void pop( ) throws underflow
• If (isempty( ))
• Throw new underflow (stackpop)
• Tos - -

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Top method for array based stack

• Return the most recently inserted item from the
  stack
• Exception underflow if the stack is empty
• Public object top( ) throws underflow
• If (isempty( ))
• Throw new underflow( stacktop)
• Return thearraytos

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Top and pop method for array based stack

• Return and remove the most recently inserted item
  from the stack
• Exception underflow if the stack is empty
• Public object topandpop( ) throws underflow
• If isempty( ) )
• Throw new underflow (stack topandpop)
• Return the arraytos - -

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Infix,postfix and prefix notations

• Consider the sum of A and B , we think of
  applying the operator to the operands A and B
  and write the sum as AB.
• This particular representation is called infix.
• There are two alternative notations for
  expressing the sum of A and B using the symbols
  A,B and , these are
• AB prefix
• AB postfix

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Suppose that we would like to rewrite ABC in
postfix

• Applying the rules of precedence,we obtained
• ABC
• A(BC) Parentheses for emphasis
• A(BC)Convert the multiplication,Let DBC
• A(D) Convert the addition
• ABC Postfix Form

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Evaluation a postfix expression (Eg ABC )

• Each operator in a postfix string refers to the
  previous two operands in the string.
• Suppose that each time we read an operand we push
  it into a stack. When we reach an operator, its
  operands will then be top two elements on the
  stack
• We can then pop these two elements, perform the
  indicated operation on them, and push the result
  on the stack.
• So that it will be available for use as an
  operand of the next operator.

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Algorithm for evaluating a postfix expression

• Initialize a stack, opndstk to be empty.
• scan the input string reading one element at a
  time into symb
• While there are more characters in the input
  string
• Do begin
• Symb next input character
• (Read symbol )

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Algorithm for evaluating a postfix expression
(Cond.)

• If symb is an operand
• Then push (opndstk,symb)
• Else symbol is an operator
• Begin
• Opnd2pop(opndstk) Opnd1pop(opndnstk)
• Value result of applying symb to opnd1 opnd2
• Push(opndstk,value)
• End else begin
• End while do begin
• Result pop (opndstk)

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Question 1 Evaluate the following expression
in postfix 623-382/23

• Final answer is
• 49
• 51
• 52
• 7
• None of these

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Evaluate- 623-382/23

• Symbol opnd1 opnd2 value opndstk
• 6 6
• 2 6,2
• 3 6,2,3
• 2 3 5 6,5
• - 6 5 1 1
• 3 6 5 1 1,3

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Evaluate- 623-382/23

• Symbol opnd1 opnd2 value opndstk
• 8 6 5 1
  1,3,8
• 2 6 5 1
  1,3,8,2
• / 8 2
  4 1,3,4
• 3 4 7 1,7
• 1 7 7 7
• 2 1 7 7
  7,2
• 7 2 49
  49
• 3 7 2 49
  49,3
• 49 3 52
  52

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Question 2

• Consider the following arithmetic expression p
  written in postfix notation
• P5,6,2,,,12,4, /, -
• (commas are used to separate the elements of p
  so that 5,6,2 is not interpreted as the number
  562

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Which of the following statements are correct ?

• I Equivalent infix is 5(62)-12/4
• ii Final stack value is 37
• iii There is a one stack intermediate value . It
  is equal to 40
• iv One of the stack intermediate value is 32
• v Equivalent infix is 95(62)-12)/4