Infix to Postfix [EASY]

Company: VMWare, Paytm

Question:

Given an infix expression. Conver the infix expression to postfix expression.

Infix expression:The expression of the form a op b. When an operator is in-between every pair of operands.

Postfix expression:The expression of the form a b op. When an operator is followed for every pair of operands.

Input:

The first line of input contains an integer T denoting the number of test cases.
The next T lines contains an infix expression.The expression contains all characters and ^,*,/,+,-.

Output:
Output the infix expression to postfix expression.

Constraints:
1<=T<=100
1<=length of expression<=30

Example:
Input:
2
a+b*(c^d-e)^(f+g*h)-i
A*(B+C)/D

Output:

abcd^e-fgh*+^*+i-
ABC+*D/

CODE:

import java.util.*;

import java.lang.*;

import java.io.*;

class Solution {

    // 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

        Stack<Character> stack = new Stack<>();

         

        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.pop();

                 

                if (!stack.isEmpty() && stack.peek() != '(')

                    return "Invalid Expression"; // invalid expression                

                else

                    stack.pop();

            }

            else // an operator is encountered

            {

                while (!stack.isEmpty() && Prec(c) <= Prec(stack.peek()))

                    result += stack.pop();

                stack.push(c);

            }

      

        }

      

        // pop all the operators from the stack

        while (!stack.isEmpty())

            result += stack.pop();

      

        return result;

    }

   

    // Driver method 

    public static void main(String[] args) 

    {   Scanner sc = new Scanner(System.in);

        int m =sc.nextInt();

        for(int mi=0;mi<m;mi++){

        String exp = sc.next();

        System.out.println(infixToPostfix(exp));

        }

    }

}

Execution Time: 0.09 secs