150.Evaluate_Reverse_Polish_Notation(Medium).md

150. Evaluate Reverse Polish Notation (Medium)

Date and Time: Jul 15, 2024, 16:53 (EST)

Link: https://leetcode.com/problems/evaluate-reverse-polish-notation/

Question:

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 <= 10^4

• tokens[i] is either an operator: "+", "-", "*", or "/", or an integer in the range [-200, 200].

KeyPoints:

Save all elements pop from tokens that are not operators, then if we encounter one operator, we pop the latest two from stack and push their results back into stack.

My Solution:

class Solution:
    def evalRPN(self, tokens: List[str]) -> int:
        stack = []
        for i in tokens:
            if i == '+' or i == '-' or i == '*' or i == '/':
                l, r = stack.pop(), stack.pop()
                if i == '+':
                    stack.append(l + r)
                elif i == '-':
                    stack.append(r - l)
                elif i == '*':
                    stack.append(l * r)
                elif i == '/':
                    stack.append(int(r / l)) 
            else:
                stack.append(int(i))
        return stack[0]

Time Complexity: $O(n)$ we loop over tokens.
Space Complexity: $O(n)$, because we create stack that may store all elements from tokens.