Prefix to Postfix Conversion
Last Updated :
23 Jan, 2024
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:
C++
#include <iostream>
#include <stack>
using namespace std;
bool isOperator(char x)
{
switch (x) {
case '+':
case '-':
case '/':
case '*':
return true;
}
return false;
}
string preToPost(string pre_exp)
{
stack<string> s;
int length = pre_exp.size();
for (int i = length - 1; i >= 0; i--)
{
if (isOperator(pre_exp[i]))
{
string op1 = s.top();
s.pop();
string op2 = s.top();
s.pop();
string temp = op1 + op2 + pre_exp[i];
s.push(temp);
}
else {
s.push(string(1, pre_exp[i]));
}
}
return s.top();
}
int main()
{
string pre_exp = "*-A/BC-/AKL";
cout << "Postfix : " << preToPost(pre_exp);
return 0;
}
|
Java
import java.util.*;
class GFG {
static boolean isOperator(char x)
{
switch (x) {
case '+':
case '-':
case '/':
case '*':
return true;
}
return false;
}
static String preToPost(String pre_exp)
{
Stack<String> s = new Stack<String>();
int length = pre_exp.length();
for (int i = length - 1; i >= 0; i--)
{
if (isOperator(pre_exp.charAt(i)))
{
String op1 = s.peek();
s.pop();
String op2 = s.peek();
s.pop();
String temp = op1 + op2 + pre_exp.charAt(i);
s.push(temp);
}
else {
s.push(pre_exp.charAt(i) + "");
}
}
return s.peek();
}
public static void main(String args[])
{
String pre_exp = "*-A/BC-/AKL";
System.out.println("Postfix : "
+ preToPost(pre_exp));
}
}
|
Python 3
s = "*-A/BC-/AKL"
stack = []
operators = set(['+', '-', '*', '/', '^'])
s = s[::-1]
for i in s:
if i in operators:
a = stack.pop()
b = stack.pop()
temp = a+b+i
stack.append(temp)
else:
stack.append(i)
print(*stack)
|
C#
using System;
using System.Collections.Generic;
class GFG {
static bool isOperator(char x)
{
switch (x) {
case '+':
case '-':
case '/':
case '*':
return true;
}
return false;
}
static String preToPost(String pre_exp)
{
Stack<String> s = new Stack<String>();
int length = pre_exp.Length;
for (int i = length - 1; i >= 0; i--)
{
if (isOperator(pre_exp[i]))
{
String op1 = s.Peek();
s.Pop();
String op2 = s.Peek();
s.Pop();
String temp = op1 + op2 + pre_exp[i];
s.Push(temp);
}
else {
s.Push(pre_exp[i] + "");
}
}
return s.Peek();
}
public static void Main(String[] args)
{
String pre_exp = "*-A/BC-/AKL";
Console.WriteLine("Postfix : "
+ preToPost(pre_exp));
}
}
|
Javascript
<script>
function isOperator(x)
{
switch (x) {
case '+':
case '-':
case '/':
case '*':
return true;
}
return false;
}
function preToPost(pre_exp)
{
let s = [];
let length = pre_exp.length;
for (let i = length - 1; i >= 0; i--)
{
if (isOperator(pre_exp[i]))
{
let op1 = s[s.length - 1];
s.pop();
let op2 = s[s.length - 1];
s.pop();
let temp = op1 + op2 + pre_exp[i];
s.push(temp);
}
else {
s.push(pre_exp[i] + "");
}
}
return s[s.length - 1];
}
let pre_exp = "*-A/BC-/AKL";
document.write("Postfix : " + preToPost(pre_exp));
</script>
|
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.
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