Infix : An expression is called the Infix expression if the operator appears in between the operands in the expression. Simply of the form (operand1 operator operand2).
Example : (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 Prefix expression, convert it into a Infix expression.
Computers usually does the computation in either prefix or postfix (usually postfix). But for humans, its easier to understand an Infix expression rather than a prefix. Hence conversion is need for human understanding.
Examples:
Input : Prefix : *+AB-CD
Output : Infix : ((A+B)*(C-D))
Input : Prefix : *-A/BC-/AKL
Output : Infix : ((A-(B/C))*((A/K)-L))
Algorithm for Prefix to Infix:
- 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 between them.
string = (operand1 + operator + operand2)
And push the resultant string back to Stack
- Repeat the above steps until the end of Prefix expression.
- At the end stack will have only 1 string i.e resultant string
Implementation:
C++
#include <iostream>
#include <stack>
using namespace std;
bool isOperator(char x) {
switch (x) {
case '+':
case '-':
case '/':
case '*':
case '^':
case '%':
return true;
}
return false;
}
string preToInfix(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 + pre_exp[i] + op2 + ")";
s.push(temp);
}
else {
s.push(string(1, pre_exp[i]));
}
}
return s.top();
}
int main() {
string pre_exp = "*-A/BC-/AKL";
cout << "Infix : " << preToInfix(pre_exp);
return 0;
}
|
Java
import java.util.Stack;
class GFG{
static boolean isOperator(char x)
{
switch(x)
{
case '+':
case '-':
case '*':
case '/':
case '^':
case '%':
return true;
}
return false;
}
public static String convert(String str)
{
Stack<String> stack = new Stack<>();
int l = str.length();
for(int i = l - 1; i >= 0; i--)
{
char c = str.charAt(i);
if (isOperator(c))
{
String op1 = stack.pop();
String op2 = stack.pop();
String temp = "(" + op1 + c + op2 + ")";
stack.push(temp);
}
else
{
stack.push(c + "");
}
}
return stack.pop();
}
public static void main(String[] args)
{
String exp = "*-A/BC-/AKL";
System.out.println("Infix : " + convert(exp));
}
}
|
Python3
def prefixToInfix(prefix):
stack = []
i = len(prefix) - 1
while i >= 0:
if not isOperator(prefix[i]):
stack.append(prefix[i])
i -= 1
else:
str = "(" + stack.pop() + prefix[i] + stack.pop() + ")"
stack.append(str)
i -= 1
return stack.pop()
def isOperator(c):
if c == "*" or c == "+" or c == "-" or c == "/" or c == "^" or c == "(" or c == ")":
return True
else:
return False
if __name__=="__main__":
str = "*-A/BC-/AKL"
print(prefixToInfix(str))
|
C#
using System;
using System.Collections;
class GFG{
static bool isOperator(char x)
{
switch(x)
{
case '+':
case '-':
case '*':
case '/':
case '^':
case '%':
return true;
}
return false;
}
public static string convert(string str)
{
Stack stack = new Stack();
int l = str.Length;
for(int i = l - 1; i >= 0; i--)
{
char c = str[i];
if (isOperator(c))
{
string op1 = (string)stack.Pop();
string op2 = (string)stack.Pop();
string temp = "(" + op1 + c + op2 + ")";
stack.Push(temp);
}
else
{
stack.Push(c + "");
}
}
return (string)stack.Pop();
}
public static void Main(string[] args)
{
string exp = "*-A/BC-/AKL";
Console.Write("Infix : " + convert(exp));
}
}
|
Javascript
<script>
function isOperator(x)
{
switch(x)
{
case '+':
case '-':
case '*':
case '/':
case '^':
case '%':
return true;
}
return false;
}
function convert(str)
{
let stack = [];
let l = str.length;
for(let i = l - 1; i >= 0; i--)
{
let c = str[i];
if (isOperator(c))
{
let op1 = stack[stack.length - 1];
stack.pop()
let op2 = stack[stack.length - 1];
stack.pop()
let temp = "(" + op1 + c + op2 + ")";
stack.push(temp);
}
else
{
stack.push(c + "");
}
}
return stack[stack.length - 1];
}
let exp = "*-A/BC-/AKL";
document.write("Infix : " + convert(exp));
</script>
|
Output
Infix : ((A-(B/C))*((A/K)-L))
Time Complexity: O(n)
Auxiliary Space: O(n)
Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!