Program to convert infix to postfix expression in C++

Parentheses are frequently employed in mathematical formulas to make their interpretation easier to understand. However, with computers, parenthesis in an expression might lengthen the time it takes to find a solution. Various notations for describing operators and operands in an expression have been proposed to reduce computer complexity.

Let us look at a few of those notations in depth in this article, notably infix and postfix notation, as well as an example of each. Along with c++, we'll dive deep into the mechanism for converting infix to postfix notation. So, let's get this party started!

Defining Infix Notation :

Infix notation refers to the placement of the operator between its operands. It's not necessary for the operand to be a constant or a variable, it might be an expression.

Example : (p + q) * (r + s)

Three operators are used in the above phrase. The first plus operator's operands are p and q, whereas the second plus operator's operands are r and s. The outcome of these operations must be evaluated using a set of rules. Finally, after applying the addition operator to (p + q) and (r + s), the multiplication operations will be used to arrive at the final result.

Syntax :

<operator> <operator> <operator>

Remember that if an expression has just one operator, we do not need to evaluate it according to any rules. If the phrase seems to have more than only one set of operators, you may use the table below to determine the operator's priority and assess the final result.

Operators	Symbols
Addition, Subtraction	+, -
Multiplication, Division	*, /
Exponents	^
Parenthesis	( ), { }, [ ]

Table for operators and symbols

Defining Postfix Notations :

Postfix expressions, also known as reverse polish notation, are expressions wherein the operator is written after the operands.

Example : the infix phrase (p + q) may be expressed as pq+ in postfix notation.

An arithmetic expression with operators executed from left to right is known as a postfix expression. Unlike infix expressions, postfix expressions do not require the use of parentheses. Furthermore, no operator precedence or associativity rules are required, removing the requirement for programmers to learn a set of rules to aid in determining the sequence in which operations will be done.

Postfix notation algorithm :

1. Expression is scanned from the left side to right.
2. Push the operand into the stack if it is detected when scanning the expression from left to right.
3. If the operator is detected, pop the operand off the stack and conduct the computation.
4. Repeat the process, retaining the final value in the stack.

Conversion of Infix to Postfix expression using the Stack Data Structure :

The ideal way for transforming an infix expression to a postfix expression is to use the stack data structure. It retains operators until both operands are processed, then flips the sequence of operators in the postfix expression to mimic the operation order.

1. Begin by scanning the expressions from the left to the right.
2. If the scanned character is an operand, print it.
3. Else
   1. If the scanned operator's precedence is greater than the precedence of the stack's operator(or the stack is empty or has'('), then the stack's push operator is used.
   1. Otherwise, pop all operators with a higher or equal priority than the scanned operator. Push this scanning operator once you've popped them. (If a parenthesis appears during popping, halt and place the scanned operator on the stack.)
4. Push the scanned character to the stack if it is a '('.
5. If the scanned character is a ')', pop the stack and output that till another '(' appears, then discard both parentheses.
6. Repeat steps 2–6 until the whole infix, i.e. all characters, has been scanned.
7. Output for printing
8. Continue to pop and print until the stack is not empty.

With the help of the example below, we can learn how to convert infix to postfix notation using stack.

Example :

A + B - C*D + (E^F) * M/K/L * J + G is the infix expression

Let's try out the above infix expression and see what the postfix expression is.

From left to right, the above text is parsed. The components in the stack as well as the appropriate postfix expression up to that point are displayed in the table below for each token:

Steps	Element	Stack contents	Postfix Expression
1	A		A
2	+	+
3	B	+	A B
4	-	-	A B +
5	C	-	A B + C
6	*	-*	A B + C
7	D	-*	A B + C D
8	+	+	A B + C D * -
9	(	+ (	A B + C D * -
10	E	+ ( ^	A B + C D * - E
11	^	+ ( ^	A B + C D * - E
12	F	+ ( ^	A B + C D * - E F
13	)	+	A B + C D * - E F ^
14	*	+ *	A B + C D* - E F ^
15	M	+ *	A B + C D* - E F ^ M
16	/	+ /	A B + C D* - E F ^ M *
17	K	+ /	A B + C D* - E F ^M * K
18	/	+ /	A B + C D* - E F ^M * K /
19	L	+ /	A B + C D* - E F ^ M * K / L
20	*	+ *	A B + C D* - E F ^M * K / L /
21	J	+ *	A B + C D* - E F ^M * K / L / J
22	+	+	A B + C D* - E F ^M * K / L / J * +
23	G	+	A B + C D* - E F ^M * K / L / J * + G
24			A B + C D* - E F ^M * K / L / J * + G+

Explanation :

As we have an operand (A) at step 1, we append it to our postfix operation. Then we come across an operator (+) and make sure our stack is empty. The operator is pushed into our stack. Then we come across another operand (L), which we attach to our postfix operation. Our postfix expression now contains AB, and our stack now contains + after step 3. The very next element is (-), and we double-check that the top of the stack includes +, which has the same priority as (-). As a result, we add (+) to our phrase, which now appears like AB+. Because the stack is empty, (-) is placed into it. This process is repeated until the very last element of our infix expression is reached. Refer to the rules for conversion outlined in the preceding section to understand why an element is added to the postfix expression or the stack, or popped off of the stack.

Implementation :

#include<bits/stdc++.h>
using namespace std;


// Defining a new Func to return precedence of operators
int precd(char chr) {
	if (chr == '^')
		return 3;
	else if (chr == '/' || chr == '*')
		return 2;
	else if (chr == '+' || chr == '-')
		return 1;
	else
		return -1;
}


// Defining a Func for conversion of infix expression to postfix expression


string infixToPostfix(string strg) {


	stack<char> str; 


	string answr = "";


	for (int i = 0; i < strg.length(); i++) {
		char chr = strg[i];


// If the current character is added to the answr string.
 if it is an operand


		if ((chr >= 'a' && chr <= 'z') || (chr >= 'A' && chr <= 'Z') || (chr >= '0' && chr <= '9'))
			answr += chr;   
// Appending the current character of string answr
// push it to the stack, If the current character of string is an '(',.
		else if (chr == '(')
			str.push('(');




// Append the top character of stack in our answer string, If the current character of string is an ')',


// and keep on popping the top character from the stack until an '(' is encountered.
		else if (chr == ')') {
			while (str.top() != '(')
			{
				answr += str.top();    
// Appending the top character of stack in the answer
				str.pop();
			}
			str.pop();
		}


//If an operator is done being scanned
		else {
			while (!str.empty() && precd(strg[i]) <= precd(str.top())) {
				ans += str.top();
				str.pop();
			}
			str.push(chr);             
// Pushing the current character of the string in the stack
		}
	}


// Popping all the remaining elements from the stack
	while (!str.empty()) {
		answr += str.top();
		str.pop();
	}


	return answr;
}


int main() {
	string strg;
	cin >> strg;
	cout << infixToPostfix(strg);
	return 0;
}

Output :

u*v+(w-x)+y
uv*wx-+y+

Time Complexity :

The above approach to convert infix to postfix notation has a time complexity of O(n), where n is the length of the infix statement. Similarly, the conversion has an O(n) space complexity since the stack data structure reGuires the same amount of space to execute the solution.

Why is the expression needed to be represented as a postfix?

• Humans can understand and solve infix expressions because the order of operators is readily distinct, but the compiler does not have an integrated order of operators.
• As a result, to solve an Infix Expression, the compiler must scan the expression numerous times in resolving the sub-expressions in an ordered manner, which is inefficient.
• Infix expressions are changed to Postfix expressions before being evaluated to avoid this traversal.

Postfix expression’s advantages over infix expression :

• Any formula can be represented without parenthesis in postfix.
• It comes in handy when assessing formulae on machines that have stacks.
• Priority is given to infix operators.

Conclusion

Infix expressions are how we generally solve problems as humans. Computers, on the other hand, need a stack to resolve expressions. It is simple for computers to answer eGuations using prefix and postfix notation without taking into account the operator's precedence. We looked at infix and postfix notation in depth in this article, as well as the simplest way to convert infix to postfix notation using the stack data structure. To enable your programming straightforward and efficient, it is strongly advised that you properly comprehend this problem.