Given a Prefix expression, convert it into a Postfix expression.
Conversion of Prefix expression directly to Postfix without going through the process of converting them first to Infix and then to Postfix is much better in terms of computation and better understanding the expression (Computers evaluate using Postfix expression).
let's discuss about Prefix and Postfix notation:
Prefix: An expression is called the prefix expression if the operator appears in the expression before the operands. Simply of the form (operator operand1 operand2).
Example : *+AB-CD (Infix : (A+B) * (C-D) )
Postfix: An expression is called the postfix expression if the operator appears in the expression after the operands. Simply of the form (operand1 operand2 operator).
Example : AB+CD-* (Infix : (A+B * (C-D) )
Note : Follow the link for prefix to postfix online convertor.
Examples:
Input : Prefix : *+AB-CD
Output : Postfix : AB+CD-*
Explanation : Prefix to Infix : (A+B) * (C-D)
Infix to Postfix : AB+CD-*
Input : Prefix : *-A/BC-/AKL
Output : Postfix : ABC/-AK/L-*
Explanation : Prefix to Infix : (A-(B/C))*((A/K)-L)
Infix to Postfix : ABC/-AK/L-*
Algorithm for Prefix to Postfix:
- Read the Prefix expression in reverse order (from right to left)
- If the symbol is an operand, then push it onto the Stack
- If the symbol is an operator, then pop two operands from the Stack
Create a string by concatenating the two operands and the operator after them.
string = operand1 + operand2 + operator
And push the resultant string back to Stack - Repeat the above steps until end of Prefix expression.
Code for Prefix to postfix conversion:
// CPP Program to convert prefix to postfix
#include <iostream>
#include <stack>
using namespace std;
// function to check if character is operator or not
bool isOperator(char x)
{
switch (x) {
case '+':
case '-':
case '/':
case '*':
return true;
}
return false;
}
// Convert prefix to Postfix expression
string preToPost(string pre_exp)
{
stack<string> s;
// length of expression
int length = pre_exp.size();
// reading from right to left
for (int i = length - 1; i >= 0; i--)
{
// check if symbol is operator
if (isOperator(pre_exp[i]))
{
// pop two operands from stack
string op1 = s.top();
s.pop();
string op2 = s.top();
s.pop();
// concat the operands and operator
string temp = op1 + op2 + pre_exp[i];
// Push string temp back to stack
s.push(temp);
}
// if symbol is an operand
else {
// push the operand to the stack
s.push(string(1, pre_exp[i]));
}
}
// stack contains only the Postfix expression
return s.top();
}
// Driver Code
int main()
{
string pre_exp = "*-A/BC-/AKL";
cout << "Postfix : " << preToPost(pre_exp);
return 0;
}
// JavaProgram to convert prefix to postfix
import java.util.*;
class GFG {
// function to check if character
// is operator or not
static boolean isOperator(char x)
{
switch (x) {
case '+':
case '-':
case '/':
case '*':
return true;
}
return false;
}
// Convert prefix to Postfix expression
static String preToPost(String pre_exp)
{
Stack<String> s = new Stack<String>();
// length of expression
int length = pre_exp.length();
// reading from right to left
for (int i = length - 1; i >= 0; i--)
{
// check if symbol is operator
if (isOperator(pre_exp.charAt(i)))
{
// pop two operands from stack
String op1 = s.peek();
s.pop();
String op2 = s.peek();
s.pop();
// concat the operands and operator
String temp = op1 + op2 + pre_exp.charAt(i);
// Push String temp back to stack
s.push(temp);
}
// if symbol is an operand
else {
// push the operand to the stack
s.push(pre_exp.charAt(i) + "");
}
}
// stack contains only the Postfix expression
return s.peek();
}
// Driver Code
public static void main(String args[])
{
String pre_exp = "*-A/BC-/AKL";
System.out.println("Postfix : "
+ preToPost(pre_exp));
}
}
// This code is contributed by Arnab Kundu
# Write Python3 code here
# -*- coding: utf-8 -*-
# Example Input
s = "*-A/BC-/AKL"
# Stack for storing operands
stack = []
operators = set(['+', '-', '*', '/', '^'])
# Reversing the order
s = s[::-1]
# iterating through individual tokens
for i in s:
# if token is operator
if i in operators:
# pop 2 elements from stack
a = stack.pop()
b = stack.pop()
# concatenate them as operand1 +
# operand2 + operator
temp = a+b+i
stack.append(temp)
# else if operand
else:
stack.append(i)
# printing final output
print(*stack)
// C# Program to convert prefix to postfix
using System;
using System.Collections.Generic;
class GFG {
// function to check if character
// is operator or not
static bool isOperator(char x)
{
switch (x) {
case '+':
case '-':
case '/':
case '*':
return true;
}
return false;
}
// Convert prefix to Postfix expression
static String preToPost(String pre_exp)
{
Stack<String> s = new Stack<String>();
// length of expression
int length = pre_exp.Length;
// reading from right to left
for (int i = length - 1; i >= 0; i--)
{
// check if symbol is operator
if (isOperator(pre_exp[i]))
{
// pop two operands from stack
String op1 = s.Peek();
s.Pop();
String op2 = s.Peek();
s.Pop();
// concat the operands and operator
String temp = op1 + op2 + pre_exp[i];
// Push String temp back to stack
s.Push(temp);
}
// if symbol is an operand
else {
// push the operand to the stack
s.Push(pre_exp[i] + "");
}
}
// stack contains only the Postfix expression
return s.Peek();
}
// Driver Code
public static void Main(String[] args)
{
String pre_exp = "*-A/BC-/AKL";
Console.WriteLine("Postfix : "
+ preToPost(pre_exp));
}
}
/* This code contributed by PrinciRaj1992 */
// Example Input
let s = "*-A/BC-/AKL";
// Stack for storing operands
let stack = [];
const operators = new Set(['+', '-', '*', '/', '^']);
// Reversing the order
s = s.split('').reverse().join('');
// iterating through individual tokens
for (let i of s) {
// if token is operator
if (operators.has(i)) {
// pop 2 elements from stack
let a = stack.pop();
let b = stack.pop();
// concatenate them as operand1 + operand2 + operator
let temp = a + b + i;
stack.push(temp);
} else {
// else if operand
stack.push(i);
}
}
// printing final output
console.log(...stack);
Output
Postfix : ABC/-AK/L-*
Time Complexity: O(N), as we are using a loop for traversing the expression.
Auxiliary Space: O(N), as we are using stack for extra space.