Evaluate_Reverse_Polish_Notation.py

# Evaluate the value of an arithmetic expression in Reverse Polish Notation.

# Valid operators are +, -, *, and /. Each operand may be an integer or another expression.

# Note that division between two integers should truncate toward zero.

# It is guaranteed that the given RPN expression is always valid. That means the expression would always evaluate to a result, 
# and there will not be any division by zero operation.

 

# Example 1:

# Input: tokens = ["2","1","+","3","*"]
# Output: 9
# Explanation: ((2 + 1) * 3) = 9
# Example 2:

# Input: tokens = ["4","13","5","/","+"]
# Output: 6
# Explanation: (4 + (13 / 5)) = 6
# Example 3:

# Input: tokens = ["10","6","9","3","+","-11","*","/","*","17","+","5","+"]
# Output: 22
# Explanation: ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
# = ((10 * (6 / (12 * -11))) + 17) + 5
# = ((10 * (6 / -132)) + 17) + 5
# = ((10 * 0) + 17) + 5
# = (0 + 17) + 5
# = 17 + 5
# = 22
 

# Constraints:

# 1 <= tokens.length <= 104
# tokens[i] is either an operator: "+", "-", "*", or "/", or an integer in the range [-200, 200].
  
class Solution:
    def evalRPN(self, tokens: List[str]) -> int:
        # we will use a stack: if we meet a number, we push it into the stack
        # if we meet an operation, we pop the last 2 numbers and apply the operation
        # then we push the result
        stack = []
        operations = ['+', '-', '*', '/']
        
        for token in tokens:
            # if the token is an operation
            if token in operations:
                # we pop two operands from stack and apply the operation
                second = int(stack.pop())
                first = int(stack.pop())
                if token == '+':
                    result = first + second
                elif token == '-':
                    result = first - second
                elif token == '*':
                    result = first * second
                else:
                    result = int(first / second)
                # now we push the result back to the stack
                stack.append(result)
            else:
                stack.append(token)
        
        # we know at this point, there is only one number in the stack
        return stack[0]