Infix to Postfix Conversion Using Stack in DSA Go - Build from Scratch

Infix to Postfix Conversion Using Stack

📖 Scenario: You are building a simple calculator that converts mathematical expressions from the common infix form (like 3 + 4) to postfix form (like 3 4 +) which computers find easier to evaluate.This conversion uses a stack to reorder operators and operands correctly.

🎯 Goal: Build a Go program that converts a given infix expression string into its postfix form using a stack.

📋 What You'll Learn

Create a string variable infix with the exact value "A+B*C-D/E"

Create a map called precedence that assigns integer precedence values to operators: '+' and '-' as 1, '*' and '/' as 2

Write a function isOperator that returns true if a character is one of '+', '-', '*', '/'

Write a function infixToPostfix that converts the infix string to postfix using a stack and the precedence map

Print the resulting postfix expression exactly as a string with no spaces

💡 Why This Matters

🌍 Real World

Converting infix expressions to postfix is a key step in calculators and compilers to simplify expression evaluation.

💼 Career

Understanding stacks and expression parsing is essential for software engineers working on interpreters, compilers, and calculators.

1

Create the infix expression string

Create a string variable called infix and set it to the exact value "A+B*C-D/E"

DSA Go

# Create the infix expression string
# Your code here

Hint

Use infix := "A+B*C-D/E" to create the string.

2

Create the operator precedence map

Create a map called precedence that assigns integer precedence values to operators: '+' and '-' as 1, '*' and '/' as 2

DSA Go

package main

func main() {
    infix := "A+B*C-D/E"
    // Create the precedence map
    // Your code here
}

Hint

Use precedence := map[rune]int{'+' : 1, '-' : 1, '*' : 2, '/' : 2}

3

Write the infixToPostfix conversion function

Write a function called isOperator that returns true if a rune is one of '+', '-', '*', '/'. Then write a function called infixToPostfix that takes the infix string and the precedence map and returns the postfix string using a stack to reorder operators correctly.

DSA Go

package main

func isOperator(c rune) bool {
    // Your code here
}

func infixToPostfix(infix string, precedence map[rune]int) string {
    // Your code here
}

func main() {
    infix := "A+B*C-D/E"
    precedence := map[rune]int{
        '+': 1,
        '-': 1,
        '*': 2,
        '/': 2,
    }
}

Hint

Use a stack slice of runes to hold operators. Append operands directly to postfix. Pop from stack while top operator has higher or equal precedence.

4

Print the postfix expression

Call the infixToPostfix function with infix and precedence and print the returned postfix string exactly.

DSA Go

package main

func isOperator(c rune) bool {
    return c == '+' || c == '-' || c == '*' || c == '/'
}

func infixToPostfix(infix string, precedence map[rune]int) string {
    stack := []rune{}
    postfix := []rune{}

    for _, ch := range infix {
        if (ch >= 'A' && ch = 'a' && ch  0 && precedence[stack[len(stack)-1]] >= precedence[ch] {
                postfix = append(postfix, stack[len(stack)-1])
                stack = stack[:len(stack)-1]
            }
            stack = append(stack, ch)
        }
    }

    for len(stack) > 0 {
        postfix = append(postfix, stack[len(stack)-1])
        stack = stack[:len(stack)-1]
    }

    return string(postfix)
}

func main() {
    infix := "A+B*C-D/E"
    precedence := map[rune]int{
        '+': 1,
        '-': 1,
        '*': 2,
        '/': 2,
    }
    // Print the postfix expression
    // Your code here
}

Hint

Call infixToPostfix(infix, precedence) and print the result using fmt.Println.

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 ')'