17CS1102 DATA STRUCTURES © 2016 KL University – The contents of this presentation are an intellectual and copyrighted property of KL University. ALL RIGHTS RESERVED

17CS1102 (Data Structures) Topic: Applications of stack Indicator:4 At the end of this session, Student will be able to: Apply stack to check if the given expression has balanced symbols or not. (balancing symbols) Evaluate Postfix Expression Convert Infix to Postfix Apply stack in function calls

Applications of Stacks Matching Balancing Parenthesis Evaluating postfix expression Infix to postfix expression Function Calls Applications of Stacks

Matching Balancing Parenthesis Given an expression of ‘(‘ and ‘)’, a ‘(‘ must match with a ‘)’, or else it is illegal. ( )( ), ( ( ( ) ) ), ( ( ( ) ) ( ) ) are all legal, while ( ( ) (, ) ( ) ( are all illegal. Obviously, counting the numbers of ‘(‘ and ‘)’ in the expression is not enough. Applications of Stacks

Matching Balancing Parenthesis Insights  By convention, the expression is reading from left-to-right. A ‘)’ is matched with the closest, unmatched ‘(‘ on its left. For example, ( ( ( ) ) ( ) ) Applications of Stacks

Matching Balancing Parenthesis Suppose we have a ‘)’. How do we know which ‘(’ is closest and unmatched? If we read the expression from left-to-right, the MOST RECENTLY UNMATCHED ‘(’ is cancelled with ‘)’. How can we keep track of the MOST RECENTLY READ (LAST) ‘(’ ? (keep in mind there are many pending unmatched ‘(’). Which data structure is keeping track of the most recent item ? Stack  LIFO structure Applications of Stacks

Matching Balancing Parenthesis When we see a ‘(‘, we push it into a stack. When we see a ‘)’, we pop ‘(‘ from the stack. This ‘(‘ matches with current ‘)’. What happen if the stack is empty (meaning there is no such ‘(’) at that moment ? What happen if we have finished reading the expression, but the stack is not empty ? It means there are more ‘(‘ than ‘)’ in the expression. It means the input is illegal. It means the input is illegal. Applications of Stacks

Postfix Expression Evaluation Infix expression Suppose the expression only has operators ‘*’ and ‘+’; and ‘*’ has a higher precedence than ‘+’. So, 5+2+3 = 10, and 1+2*4=9, etc. The expression may also have parenthesis to complicate the problem, i.e., (1+2)*4=12 (1+2*5+1)*3=36. (1+2*(5+1))*3=39. Applications of Stacks

Postfix Expression Evaluation 1 3 + 1 2 4 * + 1 2 + 4 * 6 5 2 3 + 8 * + 3 + * are all postfix expressions. No ‘(‘, ‘)’ is in the expression. To evaluate a postfix expression, we need a stack. Applications of Stacks

Postfix Expression Evaluation Evaluate a postfix expression. Read the expression from left-to-right. When a number is seen, it is pushed onto the stack. When an operator is seen, the operator is applied to the two numbers that are popped from the stack. The result is pushed on the stack. Applications of Stacks

Infix to Postfix Expressions Example : 6 5 2 3 + 8 * + 3 + * Applications of Stacks

Infix to Postfix Expressions How to convert an infix expression to a postfix expression? Use a stack. Read the infix expression from left-to-right. When an operand (number) is read, output it. If an operator (other than ‘(‘, ‘)’) is read, pop the stack (and output the operator) until a lower precedence operator or ‘(‘ is on the top of stack. Then push the current operator on the stack. If ‘(‘ is read, push it on the stack. If a ‘)‘ is read, pop the stacks (and output the operators), until we meet ‘(’. Pop that ‘(’ (do no output it). Finally, if we read the end of the expression, pop the stack (and output the operators) until the stack is empty. Applications of Stacks

Infix to Postfix Expressions Example : (1+2*(5+1))*3 postfix expression  1 2 5 1 + * + 3 * Applications of Stacks