# 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 :

- Expression is scanned from the
**left side to right**. **Push**the operand into the stack if it is detected when scanning the expression from left to right.- If the operator is detected,
**pop**the operand off the stack and conduct the computation. - 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.

- Begin by
**scanning the expressions**from the left to the right. - If the
**scanned character**is an operand, print it. - Else
- 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.

- 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.)

- If the scanned operator's precedence is greater than the
**Push**the scanned character to the stack if it is a**'('.**- If the scanned character is a
**')',****pop**the stack and output that till another '(' appears, then discard both parentheses. **Repeat steps 2–6**until the whole infix, i.e. all characters, has been scanned.- Output for printing
- 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.