Infix to Postfix Expression

Try it on GfG Practice

Write a program to convert an Infix expression to Postfix form.

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

Examples:

Input: s = "A*(B+C)/D"
Output: ABC+*D/

Input: s = "a+b*(c^d-e)^(f+g*h)-i"
Output: abcd^e-fgh*+^*+i- 

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. 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. 

Conversion of an Infix expression to Postfix expression

To convert infix expression to postfix expression, use the stack data structure. Scan the infix expression from left to right. Whenever we get an operand, add it to the postfix expression and if we get an operator or parenthesis add it to the stack by maintaining their precedence.

Below are the steps to implement the above idea:

1. Scan the infix expression from left to right. 
2. If the scanned character is an operand, put it in the postfix expression. 
3. Otherwise, do the following
   • If the precedence of the current scanned operator is higher than the precedence of the operator on top of the stack, or if the stack is empty, or if the stack contains a ‘(‘, then push the current operator onto the stack.
   • Else, pop all operators from the stack that have precedence higher than or equal to that of the current operator. After that push the current operator onto the stack.
4. If the scanned character is a ‘(‘, push it to the stack. 
5. If the scanned character is a ‘)’, pop the stack and output it until a ‘(‘is encountered, and discard both the parenthesis. 
6. Repeat steps 2-5 until the infix expression is scanned. 
7. Once the scanning is over, Pop the stack and add the operators in the postfix expression until it is not empty.
8. Finally, print the postfix expression.

Illustration:

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

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

string infixToPostfix(string s) {
    stack<char> st;
    string res;

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

        // If the scanned character is
        // an operand, add it to the output string.
        if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') || (c >= '0' && c <= '9'))
            res += 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 add to the output string from the stack
        // until an ‘(‘ is encountered.
        else if (c == ')') {
            while (st.top() != '(') {
                res += st.top();
                st.pop();
            }
            st.pop(); 
        }

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

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

    return res;
}

int main() {
    string exp = "a+b*(c^d-e)^(f+g*h)-i";
    cout << infixToPostfix(exp);
    return 0;
}

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

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

void infixToPostfix(char* exp) {
    int len = strlen(exp);
    char result[len + 1];
    char stack[len];
    int j = 0;
    int top = -1;

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

        // If the scanned character is 
        // an operand, add it to the output string.
        if (isalnum(c))
            result[j++] = c;

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

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

        // If an operator is scanned
        else {
            while (top != -1 && (prec(c) < prec(stack[top]) ||
                                 prec(c) == prec(stack[top]))) {
                result[j++] = stack[top--];
            }
            stack[++top] = c;
        }
    }

    // Pop all the remaining elements from the stack
    while (top != -1) {
        result[j++] = stack[top--];
    }

    result[j] = '\0';
    printf("%s\n", result);
}

int main() {
    char exp[] = "a+b*(c^d-e)^(f+g*h)-i";
    infixToPostfix(exp);
    return 0;
}

import java.util.Stack;

public class InfixToPostfix {
    static int prec(char c) {
        if (c == '^') return 3;
        else if (c == '/' || c == '*') return 2;
        else if (c == '+' || c == '-') return 1;
        else return -1;
    }

    static String infixToPostfix(String s) {
        Stack<Character> st = new Stack<>();
        StringBuilder res = new StringBuilder();

        for (char c : s.toCharArray()) {
            
            // If the scanned character is an operand, 
            // add it to the output string.
            if (Character.isLetterOrDigit(c)) {
                res.append(c);
            } 
            
            // If the scanned character is an ‘(‘, 
            // push it to the stack.
            else if (c == '(') {
                st.push(c);
            } 
            
            // If the scanned character is an ‘)’,
            // pop and add to the output string from the stack
            // until an ‘(‘ is encountered.
            else if (c == ')') {
                while (!st.isEmpty() && st.peek() != '(') {
                    res.append(st.pop());
                }
                st.pop();
            } 
            
            // If an operator is scanned
            else {
                while (!st.isEmpty() && prec(c) <= prec(st.peek())) {
                    res.append(st.pop());
                }
                st.push(c);
            }
        }

        // Pop all the remaining elements from the stack
        while (!st.isEmpty()) {
            res.append(st.pop());
        }

        return res.toString();
    }

    public static void main(String[] args) {
        String exp = "a+b*(c^d-e)^(f+g*h)-i";
        System.out.println(infixToPostfix(exp));
    }
}

def prec(c):
    if c == '^':
        return 3
    elif c in ('/', '*'):
        return 2
    elif c in ('+', '-'):
        return 1
    else:
        return -1

def infix_to_postfix(s):
    st = []
    res = ''

    for c in s:
        
        # If the scanned character is an operand, 
        # add it to the output string.
        if c.isalnum():
            res += c

        # If the scanned character is an ‘(‘, 
        # push it to the stack.
        elif c == '(': 
            st.append('(')

        # If the scanned character is an ‘)’,
        # pop and add to the output string from the stack
        # until an ‘(‘ is encountered.
        elif c == ')':
            while st and st[-1] != '(': 
                res += st.pop()
            st.pop()

        # If an operator is scanned
        else:
            while st and prec(c) <= prec(st[-1]):
                res += st.pop()
            st.append(c)

    # Pop all the remaining elements from the stack
    while st:
        res += st.pop()

    return res

if __name__ == '__main__':
    exp = 'a+b*(c^d-e)^(f+g*h)-i'
    print(infix_to_postfix(exp))

using System;
using System.Collections.Generic;

class GfG {
    static int Prec(char c) {
        if (c == '^') return 3;
        else if (c == '/' || c == '*') return 2;
        else if (c == '+' || c == '-') return 1;
        else return -1;
    }

    static string InfixToPostfix(string s) {
        Stack<char> st = new Stack<char>();
        string res = "";

        foreach (char c in s) {
            
            // If the scanned character is an operand, 
            // add it to the output string.
            if (char.IsLetterOrDigit(c)) {
                res += c;
            } 
            
            // If the scanned character is an ‘(‘, 
            // push it to the stack.
            else if (c == '(') {
                st.Push(c);
            } 
            
            // If the scanned character is an ‘)’,
            // pop and add to the output string from the stack
            // until an ‘(‘ is encountered.
            else if (c == ')') {
                while (st.Count > 0 && st.Peek() != '(') {
                    res += st.Pop();
                }
                st.Pop();
            } 
            
            // If an operator is scanned
            else {
                while (st.Count > 0 && Prec(c) <= Prec(st.Peek())) {
                    res += st.Pop();
                }
                st.Push(c);
            }
        }

        // Pop all the remaining elements from the stack
        while (st.Count > 0) {
            res += st.Pop();
        }

        return res;
    }

    static void Main() {
        string exp = "a+b*(c^d-e)^(f+g*h)-i";
        Console.WriteLine(InfixToPostfix(exp));
    }
}

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

function infixToPostfix(s) {
    let st = [];
    let res = '';

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

        // If the scanned character is an operand, add it to the output string.
        if (/[a-zA-Z0-9]/.test(c))
            res += 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 add to the output string from the stack
        // until an ‘(‘ is encountered.
        else if (c === ')') {
            while (st.length && st[st.length - 1] !== '(') {
                res += st.pop();
            }
            st.pop();
        }

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

    // Pop all the remaining elements from the stack
    while (st.length) {
        res += st.pop();
    }

    return res;
}

const exp = 'a+b*(c^d-e)^(f+g*h)-i';
console.log(infixToPostfix(exp));

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

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

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