Postfix to Prefix Conversion
Last Updated :
18 Apr, 2023
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) )
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) )
Given a Postfix expression, convert it into a Prefix expression.
Conversion of Postfix expression directly to Prefix without going through the process of converting them first to Infix and then to Prefix is much better in terms of computation and better understanding the expression (Computers evaluate using Postfix expression).
Examples:
Input : Postfix : AB+CD-*
Output : Prefix : *+AB-CD
Explanation : Postfix to Infix : (A+B) * (C-D)
Infix to Prefix : *+AB-CD
Input : Postfix : ABC/-AK/L-*
Output : Prefix : *-A/BC-/AKL
Explanation : Postfix to Infix : ((A-(B/C))*((A/K)-L))
Infix to Prefix : *-A/BC-/AKL
Algorithm for Postfix to Prefix:
- Read the Postfix expression from left to right
- 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 before them.
string = operator + operand2 + operand1
And push the resultant string back to Stack
- Repeat the above steps until end of Postfix expression.
Below is the implementation of the above idea:
C++
#include <bits/stdc++.h>
using namespace std;
bool isOperator(char x)
{
switch (x) {
case '+':
case '-':
case '/':
case '*':
return true;
}
return false;
}
string postToPre(string post_exp)
{
stack<string> s;
int length = post_exp.size();
for (int i = 0; i < length; i++) {
if (isOperator(post_exp[i])) {
string op1 = s.top();
s.pop();
string op2 = s.top();
s.pop();
string temp = post_exp[i] + op2 + op1;
s.push(temp);
}
else {
s.push(string(1, post_exp[i]));
}
}
string ans = "";
while (!s.empty()) {
ans += s.top();
s.pop();
}
return ans;
}
int main()
{
string post_exp = "ABC/-AK/L-*";
cout << "Prefix : " << postToPre(post_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 postToPre(String post_exp)
{
Stack<String> s = new Stack<String>();
int length = post_exp.length();
for (int i = 0; i < length; i++) {
if (isOperator(post_exp.charAt(i))) {
String op1 = s.peek();
s.pop();
String op2 = s.peek();
s.pop();
String temp
= post_exp.charAt(i) + op2 + op1;
s.push(temp);
}
else {
s.push(post_exp.charAt(i) + "");
}
}
String ans = "";
for (String i : s)
ans += i;
return ans;
}
public static void main(String args[])
{
String post_exp = "ABC/-AK/L-*";
System.out.println("Prefix : "
+ postToPre(post_exp));
}
}
|
Python3
def isOperator(x):
if x == "+":
return True
if x == "-":
return True
if x == "/":
return True
if x == "*":
return True
return False
def postToPre(post_exp):
s = []
length = len(post_exp)
for i in range(length):
if (isOperator(post_exp[i])):
op1 = s[-1]
s.pop()
op2 = s[-1]
s.pop()
temp = post_exp[i] + op2 + op1
s.append(temp)
else:
s.append(post_exp[i])
ans = ""
for i in s:
ans += i
return ans
if __name__ == "__main__":
post_exp = "AB+CD-"
print("Prefix : ", postToPre(post_exp))
|
C#
using System;
using System.Collections;
class GFG {
static Boolean isOperator(char x)
{
switch (x) {
case '+':
case '-':
case '/':
case '*':
return true;
}
return false;
}
static String postToPre(String post_exp)
{
Stack s = new Stack();
int length = post_exp.Length;
for (int i = 0; i < length; i++) {
if (isOperator(post_exp[i])) {
String op1 = (String)s.Peek();
s.Pop();
String op2 = (String)s.Peek();
s.Pop();
String temp = post_exp[i] + op2 + op1;
s.Push(temp);
}
else {
s.Push(post_exp[i] + "");
}
}
String ans = "";
while (s.Count > 0)
ans += s.Pop();
return ans;
}
public static void Main(String[] args)
{
String post_exp = "ABC/-AK/L-*";
Console.WriteLine("Prefix : "
+ postToPre(post_exp));
}
}
|
Javascript
<script>
function isOperator(x)
{
switch (x) {
case '+':
case '-':
case '/':
case '*':
return true;
}
return false;
}
function postToPre(post_exp)
{
let s = [];
let length = post_exp.length;
for (let i = 0; i < length; i++) {
if (isOperator(post_exp[i])) {
let op1 = s[s.length - 1];
s.pop();
let op2 = s[s.length - 1];
s.pop();
let temp = post_exp[i] + op2 + op1;
s.push(temp);
}
else {
s.push(post_exp[i] + "");
}
}
let ans = "";
while (s.length > 0)
ans += s.pop();
return ans;
}
let post_exp = "ABC/-AK/L-*";
document.write("Prefix : " + postToPre(post_exp));
</script>
|
Output
Prefix : *-A/BC-/AKL
Time Complexity: O(N) // In the above-given approach, there is one loop for iterating over string which takes O(N) time in worst case. Therefore, the time complexity for this approach will be O(N).
Auxiliary Space: O(N) // we are using an empty stack as well as empty string to store the expression hence space taken is linear
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