Prefix to Infix Conversion

12 Jun, 2022
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
C++
Java
Python3
C#
Javascript
// C++ Program to convert prefix to Infix#include <iostream>#include <stack>using namespace std; // function to check if character is operator or notbool isOperator(char x) {  switch (x) {  case '+':  case '-':  case '/':  case '*':  case '^':  case '%':    return true;  }  return false;} // Convert prefix to Infix expressionstring preToInfix(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 + pre_exp[i] + op2 + ")";       // 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 now contains the Infix expression  return s.top();} // Driver Codeint main() {  string pre_exp = "*-A/BC-/AKL";  cout << "Infix : " << preToInfix(pre_exp);  return 0;}
// Java program to convert prefix to Infiximport java.util.Stack; class GFG{ // Function to check if character// is operator or not    static    boolean isOperator(char x){    switch(x)    {        case '+':        case '-':        case '*':        case '/':        case '^':        case '%':            return true;    }    return false;} // Convert prefix to Infix expressionpublic static String convert(String str){    Stack<String> stack = new Stack<>();         // Length of expression    int l = str.length();         // Reading from right to left    for(int i = l - 1; i >= 0; i--)    {        char c = str.charAt(i);        if (isOperator(c))        {            String op1 = stack.pop();            String op2 = stack.pop();                         // Concat the operands and operator            String temp = "(" + op1 + c + op2 + ")";            stack.push(temp);        }        else        {                         // To make character to string            stack.push(c + "");        }    }    return stack.pop();} // Driver codepublic static void main(String[] args){    String exp = "*-A/BC-/AKL";    System.out.println("Infix : " + convert(exp));}} // This code is contributed by abbeyme
# Python Program to convert prefix to Infixdef prefixToInfix(prefix):    stack = []         # read prefix in reverse order    i = len(prefix) - 1    while i >= 0:        if not isOperator(prefix[i]):                         # symbol is operand            stack.append(prefix[i])            i -= 1        else:                       # symbol is operator            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 # Driver codeif __name__=="__main__":    str = "*-A/BC-/AKL"    print(prefixToInfix(str))     # This code is contributed by avishekarora
// C# program to convert prefix to Infixusing System;using System.Collections; class GFG{  // Function to check if character// is operator or not    static bool isOperator(char x){    switch(x)    {        case '+':        case '-':        case '*':        case '/':        case '^':        case '%':            return true;    }    return false;}  // Convert prefix to Infix expressionpublic static string convert(string str){    Stack stack = new Stack();          // Length of expression    int l = str.Length;          // Reading from right to left    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();                          // Concat the operands and operator            string temp = "(" + op1 + c + op2 + ")";            stack.Push(temp);        }        else        {                         // To make character to string            stack.Push(c + "");        }    }    return (string)stack.Pop();}  // Driver codepublic static void Main(string[] args){    string exp = "*-A/BC-/AKL";         Console.Write("Infix : " + convert(exp));}} // This code is contributed by rutvik_56
<script>    // Javascript program to convert prefix to Infix         // Function to check if character    // is operator or not       function isOperator(x)    {        switch(x)        {            case '+':            case '-':            case '*':            case '/':            case '^':            case '%':                return true;        }        return false;    }     // Convert prefix to Infix expression    function convert(str)    {        let stack = [];         // Length of expression        let l = str.length;         // Reading from right to left        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()                 // Concat the operands and operator                let temp = "(" + op1 + c + op2 + ")";                stack.push(temp);            }            else            {                 // To make character to string                stack.push(c + "");            }        }        return stack[stack.length - 1];    }         let exp = "*-A/BC-/AKL";          document.write("Infix : " + convert(exp));         // This code is contributed by suresh07.</script>
Infix : ((A-(B/C))*((A/K)-L))
Time Complexity: O(n)
Auxiliary Space: O(n)
abhy1209120
rutvik_56
avishekarora
suresh07
swatijha0829
ajaymakvana
expression-evaluation
Stack
Strings
Strings
Stack
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