Introduction

In this blog post, we will explore how to solve the LeetCode problem "150" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an array of strings tokens that represents an arithmetic expression in a Reverse Polish Notation. Evaluate the expression. Return an integer that represents the value of the expression. Note that: The valid operators are '+', '-', '', and '/'. Each operand may be an integer or another expression. The division between two integers always truncates toward zero. There will not be any division by zero. The input represents a valid arithmetic expression in a reverse polish notation. The answer and all the intermediate calculations can be represented in a 32-bit integer. 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].

Explanation

• Stack-Based Evaluation: Use a stack to store operands. When an operator is encountered, pop the necessary number of operands from the stack, perform the operation, and push the result back onto the stack.
  • Operator Handling: Implement a mapping or conditional logic to correctly apply the appropriate arithmetic operation based on the operator token.
  • Integer Conversion: Convert the string tokens representing integers into actual integer values for computation.

• Runtime Complexity: O(n), where n is the number of tokens. Storage Complexity: O(n) in the worst case (e.g., all operands).

Code

    class Solution:
    def evalRPN(self, tokens: list[str]) -> int:
        stack = []
        operators = {
            "+": lambda x, y: x + y,
            "-": lambda x, y: x - y,
            "*": lambda x, y: x * y,
            "/": lambda x, y: int(x / y),  # Truncate towards zero
        }

        for token in tokens:
            if token in operators:
                operand2 = stack.pop()
                operand1 = stack.pop()
                result = operators[token](operand1, operand2)
                stack.append(result)
            else:
                stack.append(int(token))

        return stack[0]