Convert Infix To Prefix Notation

While we use infix expressions in our day to day lives. Computers have trouble understanding this format because they need to keep in mind rules of operator precedence and also brackets. Prefix and Postfix expressions are easier for a computer to understand and evaluate.

Given two operands a and b and an operator \odot, the infix notation implies that O will be placed in between a and b i.e a \odot b . When the operator is placed after both operands i.e ab\odot, it is called postfix notation. And when the operator is placed before the operands i.e \odot a b, the expression in prefix notation.

Given any infix expression we can obtain the equivalent prefix and postfix format.

Examples:



Input : A * B + C / D
Output : + * A B/ C D 

Input : (A - B/C) * (A/K-L)
Output : *-A/BC-/AKL

To convert an infix to postfix expression refer to this article Stack | Set 2 (Infix to Postfix). We use the same to convert Infix to Prefix.

  • Step 1: Reverse the infix expression i.e A+B*C will become C*B+A. Note while reversing each ‘(‘ will become ‘)’ and each ‘)’ becomes ‘(‘.
  • Step 2: Obtain the postfix expression of the modified expression i.e CB*A+.
  • Step 3: Reverse the postfix expression. Hence in our example prefix is +A*BC.

Below is the C++ implementation of the algorithm.

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// CPP program to convert infix to prefix
#include <bits/stdc++.h>
using namespace std;
  
bool isOperator(char c)
{
    return (!isalpha(c) && !isdigit(c));
}
  
int getPriority(char C)
{
    if (C == '-' || C == '+')
        return 1;
    else if (C == '*' || C == '/')
        return 2;
    else if (C == '^')
        return 3;
    return 0;
}
  
string infixToPostfix(string infix)
{
    infix = '(' + infix + ')';
    int l = infix.size();
    stack<char> char_stack;
    string output;
  
    for (int i = 0; i < l; i++) {
  
        // If the scanned character is an 
        // operand, add it to output.
        if (isalpha(infix[i]) || isdigit(infix[i]))
            output += infix[i];
  
        // If the scanned character is an
        // ‘(‘, push it to the stack.
        else if (infix[i] == '(')
            char_stack.push('(');
  
        // If the scanned character is an
        // ‘)’, pop and output from the stack 
        // until an ‘(‘ is encountered.
        else if (infix[i] == ')') {
  
            while (char_stack.top() != '(') {
                output += char_stack.top();
                char_stack.pop();
            }
  
            // Remove '(' from the stack
            char_stack.pop(); 
        }
  
        // Operator found 
        else {
              
            if (isOperator(char_stack.top())) {
                while (getPriority(infix[i])
                   <= getPriority(char_stack.top())) {
                    output += char_stack.top();
                    char_stack.pop();
                }
  
                // Push current Operator on stack
                char_stack.push(infix[i]);
            }
        }
    }
    return output;
}
  
string infixToPrefix(string infix)
{
    /* Reverse String
     * Replace ( with ) and vice versa
     * Get Postfix
     * Reverse Postfix  *  */
    int l = infix.size();
  
    // Reverse infix
    reverse(infix.begin(), infix.end());
  
    // Replace ( with ) and vice versa
    for (int i = 0; i < l; i++) {
  
        if (infix[i] == '(') {
            infix[i] = ')';
            i++;
        }
        else if (infix[i] == ')') {
            infix[i] = '(';
            i++;
        }
    }
  
    string prefix = infixToPostfix(infix);
  
    // Reverse postfix
    reverse(prefix.begin(), prefix.end());
  
    return prefix;
}
  
// Driver code
int main()
{
    string s = ("(a-b/c)*(a/k-l)");
    cout << infixToPrefix(s) << std::endl;
    return 0;
}

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Output:


*-a/bc-/akl



Complexity:

Stack operations like push() and pop() are performed in constant time. Since we scan all the characters in the expression once the complexity is linear in time i.e  \ \mathcal{O}(n).


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Sayan Mahapatra
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