Convert Infix expression to Postfix expression

Write a program to Convert Infix expression to Postfix.

Infix expression: The expression of the form a operator b (a + b). When an operator is in-between every pair of operands.
Postfix expression: The expression of the form a b operator (ab+). When an operator is followed by every pair of operands.

Examples:

Input: A + B * C + D
Output: ABC*+D+

Input: ((A + B) – C * (D / E)) + F
Output: AB+CDE/*-F+  

Why postfix representation of the expression?

The compiler scans the expression either from left to right or from right to left. 
Consider the expression: a + b * c + d

The compiler first scans the expression to evaluate the expression b * c, then again scans the expression to add a to it. The result is then added to d after another scan. The repeated scanning makes it very inefficient and Infix expressions are easily readable and solvable by humans whereas the computer cannot differentiate the operators and parenthesis easily so, it is better to convert the expression to postfix(or prefix) form before evaluation.
The corresponding expression in postfix form is abc*+d+. The postfix expressions can be evaluated easily using a stack. 

Steps to convert Infix expression to Postfix expression using Stack:

• Scan the infix expression from left to right. 
• If the scanned character is an operand, output it. 
• Else, 
  • If the precedence and associativity of the scanned operator are greater than the precedence and associativity of the operator in the stack(or the stack is empty or the stack contains a ‘(‘ ), then push it.
  • ‘^’ operator is right associative and other operators like ‘+’,’-‘,’*’ and ‘/’ are left-associative. Check especially for a condition when both,  operator at the top of the stack and the scanned operator are ‘^’. In this condition, the precedence of the scanned operator is higher due to its right associativity. So it will be pushed into the operator stack. In all the other cases when the top of the operator stack is the same as the scanned operator, then pop the operator from the stack because of left associativity due to which the scanned operator has less precedence. 
  • Else, Pop all the operators from the stack which are greater than or equal to in precedence than that of the scanned operator. After doing that Push the scanned operator to the stack. (If you encounter parenthesis while popping then stop there and push the scanned operator in the stack.) 
• If the scanned character is an ‘(‘, push it to the stack. 
• If the scanned character is an ‘)’, pop the stack and output it until a ‘(‘ is encountered, and discard both the parenthesis. 
• Repeat steps 2-6 until the infix expression is scanned. 
• Print the output 
• Pop and output from the stack until it is not empty.

Below is the implementation of the above algorithm: 

C++14

/* C++ implementation to convert
infix expression to postfix*/

#include <bits/stdc++.h>
using namespace std;

// Function to return precedence of operators
int prec(char c)
{
    if (c == '^')
        return 3;
    else if (c == '/' || c == '*')
        return 2;
    else if (c == '+' || c == '-')
        return 1;
    else
        return -1;
}

// The main function to convert infix expression
// to postfix expression
void infixToPostfix(string s)
{

    stack<char> st; // For stack operations, we are using
                    // C++ built in stack
    string result;

    for (int i = 0; i < s.length(); i++) {
        char c = s[i];

        // If the scanned character is
        // an operand, add it to output string.
        if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z')
            || (c >= '0' && c <= '9'))
            result += c;

        // If the scanned character is an
        // ‘(‘, push it to the stack.
        else if (c == '(')
            st.push('(');

        // If the scanned character is an ‘)’,
        // pop and to output string from the stack
        // until an ‘(‘ is encountered.
        else if (c == ')') {
            while (st.top() != '(') {
                result += st.top();
                st.pop();
            }
            st.pop();
        }

        // If an operator is scanned
        else {
            while (!st.empty()
                   && prec(s[i]) <= prec(st.top())) {
                result += st.top();
                st.pop();
            }
            st.push(c);
        }
    }

    // Pop all the remaining elements from the stack
    while (!st.empty()) {
        result += st.top();
        st.pop();
    }

    cout << result << endl;
}

// Driver's code
int main()
{
    string exp = "a+b*(c^d-e)^(f+g*h)-i";

    // Function call
    infixToPostfix(exp);
    return 0;
}

C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

#define MAX_EXPR_SIZE 100

int precedence(char operator) {
    switch (operator) {
        case '+':
        case '-':
            return 1;
        case '*':
        case '/':
            return 2;
        case '^':
            return 3;
        default:
            return -1;
    }
}

int isOperator(char ch) {
    return (ch == '+' || ch == '-' || ch == '*' || ch == '/' || ch == '^');
}

char *infixToPostfix(char *infix) {
    int i, j;
    int len = strlen(infix);
    char *postfix = (char *)malloc(sizeof(char) * (len + 2));
    char stack[MAX_EXPR_SIZE];
    int top = -1;

    for (i = 0, j = 0; i < len; i++) {
        if (infix[i] == ' ' || infix[i] == '\t')
            continue;
        if (isalnum(infix[i])) {
            postfix[j++] = infix[i];
        } else if (infix[i] == '(') {
            stack[++top] = infix[i];
        } else if (infix[i] == ')') {
            while (top > -1 && stack[top] != '(')
                postfix[j++] = stack[top--];
            if (top > -1 && stack[top] != '(')
                return "Invalid Expression";
            else
                top--;
        } else if (isOperator(infix[i])) {
            while (top > -1 && precedence(stack[top]) >= precedence(infix[i]))
                postfix[j++] = stack[top--];
            stack[++top] = infix[i];
        }
    }

    while (top > -1)
        postfix[j++] = stack[top--];

    postfix[j] = '\0';
    return postfix;
}

int main() {
    char infix[MAX_EXPR_SIZE];
    printf("Enter an infix expression: ");
    gets(infix);
    char *postfix = infixToPostfix(infix);
    printf("Postfix expression: %s\n", postfix);
    free(postfix);
    return 0;
}

Java

/* Java implementation to convert
 infix expression to postfix*/
// Note that here we use ArrayDeque class for Stack
// operations

import java.util.ArrayDeque;
import java.util.Deque;
import java.util.Stack;

class Test {

    // A utility function to return
    // precedence of a given operator
    // Higher returned value means
    // higher precedence
    static int Prec(char ch)
    {
        switch (ch) {
        case '+':
        case '-':
            return 1;

        case '*':
        case '/':
            return 2;

        case '^':
            return 3;
        }
        return -1;
    }

    // The main method that converts
    // given infix expression
    // to postfix expression.
    static String infixToPostfix(String exp)
    {
        // initializing empty String for result
        String result = new String("");

        // initializing empty stack
        Deque<Character> stack
            = new ArrayDeque<Character>();

        for (int i = 0; i < exp.length(); ++i) {
            char c = exp.charAt(i);

            // If the scanned character is an
            // operand, add it to output.
            if (Character.isLetterOrDigit(c))
                result += c;

            // If the scanned character is an '(',
            // push it to the stack.
            else if (c == '(')
                stack.push(c);

            //  If the scanned character is an ')',
            // pop and output from the stack
            // until an '(' is encountered.
            else if (c == ')') {
                while (!stack.isEmpty()
                       && stack.peek() != '(') {
                    result += stack.peek();
                    stack.pop();
                }

                stack.pop();
            }
            else // an operator is encountered
            {
                while (!stack.isEmpty()
                       && Prec(c) <= Prec(stack.peek())) {

                    result += stack.peek();
                    stack.pop();
                }
                stack.push(c);
            }
        }

        // pop all the operators from the stack
        while (!stack.isEmpty()) {
            if (stack.peek() == '(')
                return "Invalid Expression";
            result += stack.peek();
            stack.pop();
        }

        return result;
    }

    // Driver's code
    public static void main(String[] args)
    {
        String exp = "a+b*(c^d-e)^(f+g*h)-i";

          // Function call
        System.out.println(infixToPostfix(exp));
    }
}

Python

# Python program to convert infix expression to postfix

# Class to convert the expression


class Conversion:

    # Constructor to initialize the class variables
    def __init__(self, capacity):
        self.top = -1
        self.capacity = capacity
        # This array is used a stack
        self.array = []
        # Precedence setting
        self.output = []
        self.precedence = {'+': 1, '-': 1, '*': 2, '/': 2, '^': 3}

    # check if the stack is empty
    def isEmpty(self):
        return True if self.top == -1 else False

    # Return the value of the top of the stack
    def peek(self):
        return self.array[-1]

    # Pop the element from the stack
    def pop(self):
        if not self.isEmpty():
            self.top -= 1
            return self.array.pop()
        else:
            return "$"

    # Push the element to the stack
    def push(self, op):
        self.top += 1
        self.array.append(op)

    # A utility function to check is the given character
    # is operand
    def isOperand(self, ch):
        return ch.isalpha()

    # Check if the precedence of operator is strictly
    # less than top of stack or not
    def notGreater(self, i):
        try:
            a = self.precedence[i]
            b = self.precedence[self.peek()]
            return True if a <= b else False
        except KeyError:
            return False

    # The main function that
    # converts given infix expression
    # to postfix expression
    def infixToPostfix(self, exp):

        # Iterate over the expression for conversion
        for i in exp:
            # If the character is an operand,
            # add it to output
            if self.isOperand(i):
                self.output.append(i)

            # If the character is an '(', push it to stack
            elif i == '(':
                self.push(i)

            # If the scanned character is an ')', pop and
            # output from the stack until and '(' is found
            elif i == ')':
                while((not self.isEmpty()) and
                      self.peek() != '('):
                    a = self.pop()
                    self.output.append(a)
                if (not self.isEmpty() and self.peek() != '('):
                    return -1
                else:
                    self.pop()

            # An operator is encountered
            else:
                while(not self.isEmpty() and self.notGreater(i)):
                    self.output.append(self.pop())
                self.push(i)

        # pop all the operator from the stack
        while not self.isEmpty():
            self.output.append(self.pop())

        print "".join(self.output)


# Driver's code
if __name__ == '__main__':
    exp = "a+b*(c^d-e)^(f+g*h)-i"
    obj = Conversion(len(exp))

    # Function call
    obj.infixToPostfix(exp)

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

using System;
using System.Collections.Generic;

/* c# implementation to convert
infix expression to postfix*/
// Note that here we use Stack
// class for Stack operations

public class Test {

    // A utility function to return
    // precedence of a given operator
    // Higher returned value means higher precedence
    internal static int Prec(char ch)
    {
        switch (ch) {
        case '+':
        case '-':
            return 1;

        case '*':
        case '/':
            return 2;

        case '^':
            return 3;
        }
        return -1;
    }

    // The main method that converts given infix expression
    // to postfix expression.
    public static string infixToPostfix(string exp)
    {
        // initializing empty String for result
        string result = "";

        // initializing empty stack
        Stack<char> stack = new Stack<char>();

        for (int i = 0; i < exp.Length; ++i) {
            char c = exp[i];

            // If the scanned character is an
            // operand, add it to output.
            if (char.IsLetterOrDigit(c)) {
                result += c;
            }

            // If the scanned character is an '(',
            // push it to the stack.
            else if (c == '(') {
                stack.Push(c);
            }

            //  If the scanned character is an ')',
            // pop and output from the stack
            // until an '(' is encountered.
            else if (c == ')') {
                while (stack.Count > 0
                       && stack.Peek() != '(') {
                    result += stack.Pop();
                }

                if (stack.Count > 0
                    && stack.Peek() != '(') {
                    return "Invalid Expression"; // invalid
                                                 // expression
                }
                else {
                    stack.Pop();
                }
            }
            else // an operator is encountered
            {
                while (stack.Count > 0
                       && Prec(c) <= Prec(stack.Peek())) {
                    result += stack.Pop();
                }
                stack.Push(c);
            }
        }

        // pop all the operators from the stack
        while (stack.Count > 0) {
            result += stack.Pop();
        }

        return result;
    }

    // Driver's code
    public static void Main(string[] args)
    {
        string exp = "a+b*(c^d-e)^(f+g*h)-i";

          // Function call
        Console.WriteLine(infixToPostfix(exp));
    }
}

// This code is contributed by Shrikant13

Javascript

    /* Javascript implementation to convert
    infix expression to postfix*/

    //Function to return precedence of operators
    function prec(c) {
        if(c == '^')
            return 3;
        else if(c == '/' || c=='*')
            return 2;
        else if(c == '+' || c == '-')
            return 1;
        else
            return -1;
    }

    // The main function to convert infix expression
    //to postfix expression
    function infixToPostfix(s) {

        let st = []; //For stack operations, we are using JavaScript built in stack
        let result = "";

        for(let i = 0; i < s.length; i++) {
            let c = s[i];

            // If the scanned character is
            // an operand, add it to output string.
            if((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') || (c >= '0' && c <= '9'))
                result += c;

            // If the scanned character is an
            // ‘(‘, push it to the stack.
            else if(c == '(')
                st.push('(');

            // If the scanned character is an ‘)’,
            // pop and to output string from the stack
            // until an ‘(‘ is encountered.
            else if(c == ')') {
                while(st[st.length - 1] != '(')
                {
                    result += st[st.length - 1];
                    st.pop();
                }
                st.pop();
            }

            //If an operator is scanned
            else {
                while(st.length != 0 && prec(s[i]) <= prec(st[st.length - 1])) {
                    result += st[st.length - 1];
                    st.pop();
                }
                st.push(c);
            }
        }

        // Pop all the remaining elements from the stack
        while(st.length != 0) {
            result += st[st.length - 1];
            st.pop();
        }

        document.write(result + "</br>");
    }

    let exp = "a+b*(c^d-e)^(f+g*h)-i";
    infixToPostfix(exp);

// This code is contributed by decode2207.

Output

abcd^e-fgh*+^*+i-

Time Complexity: O(N), where N is the size of the infix expression
Auxiliary Space: O(N)

Quiz: Stack Questions
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