Data Structures & Algorithm Analysis. Lecturer: Souad Alonazi

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1 Data Structures & Algorithm Analysis Lec(3) Stacks Lecturer: Souad Alonazi

2 What is a stack? Stores a set of elements in a particular order Stack principle: LAST IN FIRST OUT = LIFO It means: the last element inserted is the first one to be removed Example Which is the first element to pick up?

3 What are stacks Top Stack of coins Stack of books Stack of dishes Computer stack 3

4 Last In First Out A top B A top C B A top D C B A top E D C B A top D C B A

5 Stack Applications Real life Pile of books Plate trays The Towers of Hanoi More applications related to computer science Program execution stack (read more from your text) Evaluating expressions

6 Evaluation of Expressions Expression Representation Techniques : Infix Prefix Expression Postfix Expression

7 From Postfix to Answer Computer System Uses Postfix form to represent expression. The reason to convert infix to postfix expression is that we can compute the answer of postfix expression easier by using a stack. Following is the example that shows evaluation of the Postfix expression using stack as data structure.

8 From Postfix to Answer Ex: * First, push(10) into the stack 10

9 From Postfix to Answer Ex: * Then, push(2) into the stack 2 10

10 From Postfix to Answer Ex: * Push(8) into the stack

11 From Postfix to Answer Ex: * Now we see an operator *, that means we can get an new number by calculation

From Postfix to Answer Ex: 10 2 8 * + 3-8 Now we see an operator *, that means

12 From Postfix to Answer Ex: * Now we see an operator *, that means we can get an new number by calculation Pop the first two numbers * 8 = 16

13 From Postfix to Answer Ex: * Now we see an operator *, that means we can get an new number by calculation Push the new number back * 8 = 16

14 From Postfix to Answer Ex: * Then we see the next operator + and perform the calculation =

15 From Postfix to Answer Ex: * Then we see the next operator + and perform the calculation Push the new number back = 26

16 From Postfix to Answer Ex: * We see the next number 3 Push (3) into the stack 3 26

17 Compute the Answer Ex: * The last operation 26-3 = 23

18 From Postfix to Answer Ex: * The last operation = 23 answer!

19 From Postfix to Answer Algorithm: maintain a stack and scan the postfix expression from left to right If the element is a number, push it into the stack If the element is a operator O, pop twice and get A and B respectively. Calculate BOA and push it back to the stack When the expression is ended, the number in the stack is the final answer

20 Transform Infix to Postfix Now, we have to design an algorithm to transform infix expression to postfix

21 Transform Infix to Postfix Observation 1: The order of computation depends on the order of operators The parentheses must be added according to the priority of operations. The priority of operator * and / is higher then those of operation + and If there are more than one equal-priority operators, we assume that the left one s priority is higher than the right one s This is called left-to-right parsing.

22 Transform Infix to Postfix Observation 1: The order of computation depends on the order of operators (cont.) For example, to add parentheses for the expression * 8-3, we first add parenthesis to 2 * 8 since its priority is highest in the expression. Then we add parenthesis to 10 + (2 * 8) since the priorities of + and are equal, and + is on the left of -. Finally, we add parenthesis to all the expression and get ((10 + (2 * 8)) - 3).

23 Transform Infix to Postfix Observation 1: The order of computation depends on the order of operators (cont.) The computation order of expression ((10 + (2 * 8)) - 3) is: 2 * 8 = = = 23 (( ) -3) (26 3) 23

24 Transform Infix to Postfix Simplify the problem, how if there are only +/- operators?

25 Transform Infix to Postfix Simplify the problem, how if there are only +/- operators? The leftmost operator will be done first Ex:

26 Transform Infix to Postfix Simplify the problem, how if there are only +/- operators? Algorithm: maintain a stack and scan the postfix expression from left to right When we get a number, output it When we get an operator O, pop the top element in the stack if the stack is not empty and then push(o) into the stack

27 Transform Infix to Postfix Simplify the problem, how if there are only +/- operators? Algorithm: maintain a stack and scan the postfix expression from left to right When we get a number, output it When we get an operator O, pop the top element in the stack if the stack is not empty and then push(o) into the stack When the expression is ended, pop all the operators remain in the stack

28 Transform Infix to Postfix Ex: We see the first number 10, output it 10

29 Transform Infix to Postfix Ex: We see the first operator +, push(+) into the stack because at this moment the stack is empty 10

30 Transform Infix to Postfix Ex: We see the number 2, output it

31 Transform Infix to Postfix Ex: We see the operator -, pop the operator + and push(-) into the stack

32 Transform Infix to Postfix Ex: We see the number 8, output it

33 Transform Infix to Postfix Ex: We see the operator +, pop the operator - and push(+) into the stack

34 Transform Infix to Postfix Ex: We see the number 3, output it

35 Transform Infix to Postfix Ex: We come to the end of the expression, then we pop all the operators in the stack

36 Transform Infix to Postfix Ex: When we get an operator, we have to push it into the stack and pop it when we see the next operator. The reason is, we have to wait for the second operand of the operator

37 Transform Infix to Postfix How to solve the problem when there are operators +, -, *, /?

38 Transform Infix to Postfix Observation 2: scan the infix expression from left to right, if we see higherpriority operator after lower-priority one, we know that the second operand of the lower-priority operator is an expression Ex: a + b * c = a + ( b * c ) a bc* + That is, the expression b c * is the second operand of the operator +

39 Transform Infix to Postfix So, we modify the algorithm to adapt the situation

40 Transform Infix to Postfix Algorithm: maintain a stack and scan the postfix expression from left to right When we get a number, output it When we get an operator O, pop the top element in the stack until there is no operator having higher priority then O and then push(o) into the stack When the expression is ended, pop all the operators remain in the stack

41 Transform Infix to Postfix Ex: * 8-3 We see the first number 10, output it 10

42 Transform Infix to Postfix Ex: * 8-3 We see the first operator +, push it into the stack + 10

43 Transform Infix to Postfix Ex: * 8-3 We see the number 2, output it

44 Transform Infix to Postfix Ex: * 8-3 We see the operator *, since the top operator in the stack, +, has lower priority then *, push(*) *

45 Transform Infix to Postfix Ex: * 8-3 We see the number 8, output it *

46 Transform Infix to Postfix Ex: * We see the operator -, because its priority is lower then *, we pop. Also, because + is on the left of it, we pop +, too. Then we push(-) * +

47 Transform Infix to Postfix Ex: * 8-3 We see the number 3, output it * + 3

48 Transform Infix to Postfix Ex: * 8-3 Because the expression is ended, we pop all the operators in the stack * + 3 -

49 Infix user 2+3*4 a*b+5 (1+2)*7 a*b/c (a/(b-c+d))*(e-a)*c a/b-c+d*e-a*c compiler Postfix 234*+ ab* * ab*c/ abc-d+/ea-*c* ab/c-de*ac*- Postfix: no parentheses, no precedence

52 Array-based Stack Implementation Allocate an array of some size (pre-defined) Maximum N elements in stack Bottom stack element stored at element 0 last index in the array is the top Increment top when one element is pushed, decrement after pop

53 Stack Implementation: CreateS, isempty, isfull Stack creates(max_stack_size) ::= #define MAX_STACK_SIZE 100 /* maximum stack size */ typedef struct { int key; /* other fields */ } element; element stack[max_stack_size]; int top = -1; Boolean isempty(stack) ::= top< 0; Boolean isfull(stack) ::= top >= MAX_STACK_SIZE-1;

54 Push void push(int *top, element item) { /* add an item to the global stack */ if (*top >= MAX_STACK_SIZE-1) { stack_full( ); return; } stack[++*top] = item; }

55 Pop element pop(int *top) { /* return the top element from the stack */ if (*top == -1) return stack_empty( ); /* returns and error key */ return stack[(*top)--]; }

56 4-2 Stack Linked List Implementation Head Top count top Data nodes (a) Conceptual (a) Physical 56

57 4-2 Stack Linked List Implementation count top Stack head structure Stack count top end stack <integer> <node pointer> data next Stack node structure node data next end node <datatype> <node pointer> 57

58 List-based Stack Implementation: Push void push(pnode top, element item) { /* add an element to the top of the stack */ pnode temp = (pnode) malloc (sizeof (node)); if (IS_FULL(temp)) { fprintf(stderr, The memory is full\n ); exit(1); } temp->item = item; temp->next= top; top= temp; }

59 Pop element pop(pnode top) { /* delete an element from the stack */ pnode temp = top; element item; if (IS_EMPTY(temp)) { fprintf(stderr, The stack is empty\n ); exit(1); } item = temp->item; top = temp->next; free(temp); return item; }

Lecture Data Structure Stack

Lecture Data Structure Stack 1.A stack :-is an abstract Data Type (ADT), commonly used in most programming languages. It is named stack as it behaves like a real-world stack, for example a deck of cards