LC 150 - 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 <= 104
• tokens[i] is either an operator: "+", "-", "*", or "/", or an integer in the range [-200, 200].

Links

Question here and solution here

Solution

concept

Use a stack, whenever see a operator, pop the last 2 number and do the operation and append it back. If it is a number just append. In the end the stack should only have 1 number, which is the answer.

code

class Solution:
    def evalRPN(self, tokens: List[str]) -> int:
        stack = []
        for c in tokens:
            if c == "+":
                stack.append(stack.pop() + stack.pop())
            elif c == "-":
                a, b = stack.pop(), stack.pop()
                stack.append(b - a)
            elif c == "*":
                stack.append(stack.pop() * stack.pop())
            elif c == "/":
                a, b = stack.pop(), stack.pop()
                stack.append(int(float(b) / a))
            else:
                stack.append(int(c))
        return stack[0]

Complexity

time: $O(n)$
space: $O(n)$