Infix to Postfix Conversion Using Stack in DSA Go - Execution Trace

Concept Flow - Infix to Postfix Conversion Using Stack

Read next symbol from infix

↓

Is symbol operand?

Yes→Add to postfix output

No↓

Is symbol '(' ?

Yes→Push '(' on stack

No↓

Is symbol ')' ?

Yes→Pop and add to postfix until '(' found

No↓

Operator encountered

↓

While stack top has higher or equal precedence

↓

Pop from stack to postfix

↓

Push current operator on stack

↓

Repeat until infix ends

↓

Pop all remaining stack to postfix

↓

Done

This flow shows how each symbol in the infix expression is processed: operands go directly to output, '(' is pushed, ')' triggers popping until '(', and operators are pushed after popping higher precedence operators.

Execution Sample

DSA Go

infix := "A+B*C"
stack := []rune{}
postfix := ""
for _, symbol := range infix {
  // process symbol
}

Converts infix expression "A+B*C" to postfix using a stack to handle operator precedence.

Execution Table

Step	Operation	Symbol	Stack Before	Stack After	Postfix Output	Visual State
1	Read symbol	A	[]	[]	A	Postfix: A | Stack: []
2	Read symbol	+	[]	+	A	Postfix: A | Stack: [+]
3	Read symbol	B	+	+	AB	Postfix: AB | Stack: [+]
4	Read symbol	*	+	+, *	AB	Postfix: AB | Stack: [+, *]
5	Read symbol	C	+, *	+, *	ABC	Postfix: ABC | Stack: [+, *]
6	End of input		+, *		ABC*+	Pop * then + to postfix | Postfix: ABC*+ | Stack: []

💡 All symbols processed; stack emptied to postfix output.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	Final
stack	[]	[]	+	+	+, *	+, *
postfix	""	A	A	AB	AB	ABC	ABC*+
symbol	""	A	+	B	*	C	""

Key Moments - 3 Insights

Why do we push '(' directly on the stack without popping?

Because '(' marks the start of a sub-expression, we push it to know where to stop popping when we find a ')'. See step 3 in concept_flow and how ')' triggers popping until '(' in the flow.

Why do we pop operators from the stack before pushing a new operator?

We pop operators with higher or equal precedence to maintain correct order in postfix. For example, in step 4 and 5, '*' has higher precedence than '+', so '*' stays on stack until after 'C' is read.

What happens if the stack is empty when we try to pop?

If the stack is empty, no pop happens. This prevents errors and means no operators are waiting. This is shown at step 1 and 2 where stack starts empty.

Visual Quiz - 3 Questions

Look at the execution_table at Step 4. What is the stack after reading '*'?

A ['+']

B ['*']

C ['+', '*']

D []

💡 Hint

Check the 'Stack After' column at Step 4 in execution_table.

At which step does the postfix output become 'ABC'?

A Step 3

B Step 5

C Step 4

D Step 6

💡 Hint

Look at the 'Postfix Output' column in execution_table for each step.

If the input was 'A*(B+C)', what would happen when ')' is read?

A Pop until '(' is found and discard '('

B Push ')' on stack

C Add ')' to postfix output

D Ignore ')' and continue

💡 Hint

Refer to concept_flow where ')' triggers popping until '(' is found.

Concept Snapshot

Infix to Postfix Conversion Using Stack:
- Read symbols left to right
- Operands go directly to output
- '(' pushed on stack
- ')' pops until '('
- Operators pop higher/equal precedence operators before push
- At end, pop all stack to output

Full Transcript

This visual execution shows how to convert an infix expression like 'A+B*C' to postfix using a stack. Each symbol is read in order. Operands like 'A', 'B', 'C' go directly to the postfix output. Operators like '+' and '*' are pushed on the stack but before pushing, operators with higher or equal precedence are popped to maintain correct order. Parentheses '(' and ')' control sub-expressions: '(' is pushed, and ')' causes popping until '(' is found and removed. At the end, all remaining operators in the stack are popped to the postfix output. The execution table tracks each step, showing the stack and postfix output changes. Variable tracker shows how stack and output evolve. Key moments clarify common confusions about parentheses and operator precedence. The quiz tests understanding of stack state and output at different steps.

Practice

1. What is the main purpose of using a stack in Infix to Postfix Conversion?

easy

A. To temporarily hold operators and manage their precedence

B. To store the final postfix expression

C. To reverse the infix expression

D. To count the number of operands

Solution

1. Step 1: Understand the role of stack in conversion

   The stack is used to hold operators temporarily while processing the infix expression to respect operator precedence and associativity.

2. Step 2: Identify what is stored in stack

   Operands are directly added to the output, while operators are pushed to the stack to decide their order in postfix.

3. Final Answer:

   To temporarily hold operators and manage their precedence -> Option A

4. Quick Check:

   Stack holds operators for precedence [OK]

Hint: Stack holds operators, output holds operands [OK]

Common Mistakes:

• Thinking stack stores operands
• Confusing stack role with output
• Ignoring operator precedence

2. Which of the following Go code snippets correctly initializes an empty stack for infix to postfix conversion?

easy

A. var stack []rune = []rune{}

B. stack := make([]int, 0)

C. stack := []string{0}

D. var stack []int = nil

Solution

1. Step 1: Identify correct stack type for operators

   Operators are characters, so stack should be a slice of runes (characters) initialized empty.

2. Step 2: Check initialization correctness

   var stack []rune = []rune{} initializes an empty rune slice correctly. stack := make([]int, 0) uses int slice, not suitable. stack := []string{0} has syntax error (cannot use int 0 as string). var stack []int = nil is nil slice, which can cause runtime errors.

3. Final Answer:

   var stack []rune = []rune{} -> Option A

4. Quick Check:

   Empty rune slice for operator stack [OK]

Hint: Use rune slice for operator stack [OK]

Common Mistakes:

• Using int slice for operators
• Initializing with wrong type
• Using nil slice without initialization

3. What is the postfix expression output by the following Go code snippet for input "a+b*c"?

infix := "a+b*c"
// Assume correct infix to postfix conversion function called here
postfix := InfixToPostfix(infix)
fmt.Println(postfix)

medium

A. "a+bc*"

B. "ab+c*"

C. "abc*+"

D. "*abc+"

Solution

1. Step 1: Analyze operator precedence in expression

   Multiplication (*) has higher precedence than addition (+), so 'b*c' is evaluated before adding 'a'.

2. Step 2: Convert infix to postfix step-by-step

   Postfix is formed as 'a' 'b' 'c' '*' '+', which is "abc*+".

3. Final Answer:

   "abc*+" -> Option C

4. Quick Check:

   Multiplication before addition = "abc*+" [OK]

Hint: Multiply before add means * goes before + in postfix [OK]

Common Mistakes:

• Ignoring operator precedence
• Placing + before * in postfix
• Confusing operand order

4. Given this Go function snippet for infix to postfix conversion, what is the bug?

func precedence(op rune) int {
    if op == '+' || op == '-' {
        return 1
    } else if op == '*' || op == '/' {
        return 2
    }
    return 0
}

// InfixToPostfix function uses precedence(op) to compare operators
// but outputs incorrect postfix for input "a+b^c"

medium

A. Precedence function returns 0 for non-operators causing errors

B. Precedence function returns int instead of float

C. Precedence values for '+' and '-' are swapped

D. Precedence function does not handle '^' operator

Solution

1. Step 1: Check if all operators are handled

   The function handles '+', '-', '*', '/' but misses '^' (exponent) operator which is common in infix expressions.

2. Step 2: Understand impact on output

   If input contains '^', missing precedence causes wrong order in postfix output.

3. Final Answer:

   Precedence function does not handle '^' operator -> Option D

4. Quick Check:

   Missing '^' causes wrong postfix [OK]

Hint: Include all operators in precedence function [OK]

Common Mistakes:

• Ignoring missing operators
• Assuming 0 precedence is error
• Confusing precedence values

5. How would you modify the infix to postfix algorithm to correctly handle parentheses in the expression (a+b)*c?

hard

A. Ignore parentheses and treat them as operands

B. Push '(' onto stack and pop operators until '(' when ')' is found

C. Remove all parentheses before conversion

D. Treat '(' as higher precedence operator and never pop it

Solution

1. Step 1: Understand role of parentheses in infix

   Parentheses change normal precedence by grouping expressions to evaluate first.

2. Step 2: Modify stack operations for parentheses

   Push '(' onto stack when encountered. When ')' is found, pop operators until '(' is popped, discarding both parentheses.

3. Final Answer:

   Push '(' onto stack and pop operators until '(' when ')' is found -> Option B

4. Quick Check:

   Parentheses control stack push/pop [OK]

Hint: Push '(' and pop until '(' on ')' [OK]

Common Mistakes:

• Ignoring parentheses in algorithm
• Treating parentheses as operands
• Not popping until '(' on ')'