# Infix To Postfix Conversion

The concept of infix to postfix conversion involves converting an arithmetic expression written in infix notation to postfix notation. Infix notation is the commonly used mathematical notation where operators are placed between their operands.

The algorithm for converting infix to postfix notation is based on the concept of operator precedence and parentheses. We use a stack to temporarily store operators. We iterate through each element of the infix expression from left to right. If an element is an operand (number), we append it to the output. If an element is an operator, we compare its precedence with the operators on the stack. If the current operator has higher precedence, we push it onto the stack. If the current operator has lower or equal precedence, we pop operators from the stack and append them to the output until we find an operator with lower precedence or an opening parenthesis. Finally, we push the current operator onto the stack.

Let's illustrate this concept with an example in Python:

```
def infix_to_postfix(expression):
precedence = {'+': 1, '-': 1, '*': 2, '/': 2, '^': 3}
stack = []
output = []
for char in expression:
if char.isalnum():
# If the character is an operand, append it to the output
output.append(char)
elif char == '(':
# If the character is an opening parenthesis, push it onto the stack
stack.append(char)
elif char == ')':
# If the character is a closing parenthesis, pop operators from the stack and append them to the output until an opening parenthesis is found
while stack and stack[-1] != '(':
output.append(stack.pop())
# Remove the opening parenthesis from the stack
stack.pop()
else:
# If the character is an operator, compare its precedence with the operators on the stack
while stack and stack[-1] != '(' and precedence[char] <= precedence[stack[-1]]:
output.append(stack.pop())
# Push the current operator onto the stack
stack.append(char)
# Pop any remaining operators from the stack and append them to the output
while stack:
output.append(stack.pop())
# Convert the output list to a string and return it
return ''.join(output)
```

In this code, the `infix_to_postfix`

function takes an infix expression as input and returns the corresponding postfix expression. We initialize an empty stack and an empty list to store the output. We iterate through each character in the expression and perform different actions based on the type of character. If the character is an operand, we append it to the output. If the character is an opening parenthesis, we push it onto the stack. If the character is a closing parenthesis, we pop operators from the stack and append them to the output until we find an opening parenthesis. If the character is an operator, we compare its precedence with the operators on the stack and perform the necessary operations. Finally, we pop any remaining operators from the stack and append them to the output.

The time complexity of this algorithm is O(n), where n is the length of the input expression. We iterate through each character once, and each operation (push, pop, append) takes constant time. The space complexity is also O(n) because we use a stack and a list to store the operators and the output, respectively.

Overall, the infix to postfix conversion algorithm is a fundamental concept in computer science and is widely used in parsing and evaluating arithmetic expressions. It allows for easier evaluation of expressions using stack-based algorithms like postfix evaluation.

Let's solve a question from LeetCode that involves infix to postfix conversion. The problem we'll solve is "Basic Calculator II" (LeetCode problem #227). In this problem, we need to evaluate a given arithmetic expression consisting of only non-negative integers, +, -, *, and / operators.

Here's the complete code with proper comments:

```
def calculate(expression):
# Function to convert infix expression to postfix expression
def infix_to_postfix(expression):
precedence = {'+': 1, '-': 1, '*': 2, '/': 2}
stack = []
output = []
for char in expression:
if char.isdigit():
output.append(char)
elif char == '(':
stack.append(char)
elif char == ')':
while stack and stack[-1] != '(':
output.append(stack.pop())
stack.pop()
else:
while stack and stack[-1] != '(' and precedence[char] <= precedence[stack[-1]]:
output.append(stack.pop())
stack.append(char)
while stack:
output.append(stack.pop())
return output
# Function to evaluate postfix expression
def evaluate_postfix(expression):
stack = []
for char in expression:
if char.isdigit():
stack.append(int(char))
else:
operand2 = stack.pop()
operand1 = stack.pop()
if char == '+':
stack.append(operand1 + operand2)
elif char == '-':
stack.append(operand1 - operand2)
elif char == '*':
stack.append(operand1 * operand2)
else:
stack.append(int(operand1 / operand2))
return stack.pop()
# Remove any whitespaces from the expression
expression = expression.replace(" ", "")
# Convert the infix expression to postfix expression
postfix_expr = infix_to_postfix(expression)
# Evaluate the postfix expression
result = evaluate_postfix(postfix_expr)
return result
```

In this code, we define two helper functions: `infix_to_postfix`

and `evaluate_postfix`

. The `infix_to_postfix`

function converts the given infix expression to postfix notation using the algorithm we discussed earlier. The `evaluate_postfix`

function evaluates the postfix expression using a stack and performs the arithmetic operations.

In the `calculate`

function, we first remove any whitespaces from the expression. Then, we call the `infix_to_postfix`

function to convert the infix expression to postfix. Next, we call the `evaluate_postfix`

function to evaluate the postfix expression and obtain the result.

The time complexity of this solution depends on the length of the input expression. The infix to postfix conversion and postfix evaluation both take O(n) time, where n is the length of the expression. Therefore, the overall time complexity is O(n). The space complexity is also O(n) as we use stacks to store the operators and operands during the conversion and evaluation process.

Please note that this solution assumes that the given expression is valid and contains only non-negative integers, +, -, *, and / operators. Additional checks and error handling can be added as per the problem requirements.