772. Basic Calculator III - Explanation

Description

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, '+', '-', '*', '/' operators, and open '(' and closing parentheses ')'. The integer division should truncate toward zero.

You may assume that the given expression is always valid. All intermediate results will be in the range of [-2³¹, 2³¹ - 1].

Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval().

Example 1:

Input: s = "1+1"

Output: 2

Example 2:

Input: s = "6-4/2"

Output: 4

Example 3:

Input: s = "2*(5+5*2)/3+(6/2+8)"

Output: 21

Constraints:

1 <= s <= 10⁴
s consists of digits, '+', '-', '*', '/', '(', and ')'.
s is a valid expression.

Company Tags

1. Stack

class Solution:
    def calculate(self, s: str) -> int:
        def evaluate(x, y, operator):
            if operator == "+":
                return x
            if operator == "-":
                return -x
            if operator == "*":
                return x * y
            return int(x / y)
        
        stack = []
        curr = 0
        previous_operator = "+"
        s += "@"
        
        for c in s:
            if c.isdigit():
                curr = curr * 10 + int(c)
            elif c == "(":
                stack.append(previous_operator)
                previous_operator = "+"
            else:
                if previous_operator in "*/":
                    stack.append(evaluate(stack.pop(), curr, previous_operator))
                else:
                    stack.append(evaluate(curr, 0, previous_operator))
                
                curr = 0
                previous_operator = c
                if c == ")":
                    while type(stack[-1]) == int:
                        curr += stack.pop()
                    previous_operator = stack.pop()

        return sum(stack)

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(n)$

Where $n$ is the length of the expression

2. Solve Isolated Expressions With Recursion

class Solution:
    def calculate(self, s: str) -> int:
        def evaluate(x, y, operator):
            if operator == "+":
                return x
            if operator == "-":
                return -x
            if operator == "*":
                return x * y
            return int(x / y)
        
        def solve(i):
            stack = []
            curr = 0
            previous_operator = "+"
            
            while i[0] < len(s):
                c = s[i[0]]
                if c == "(":
                    i[0] += 1
                    curr = solve(i)
                elif c.isdigit():
                    curr = curr * 10 + int(c)
                else:
                    if previous_operator in "*/":
                        stack.append(evaluate(stack.pop(), curr, previous_operator))
                    else:
                        stack.append(evaluate(curr, 0, previous_operator))
                     
                    if c == ")":
                        break
                    
                    curr = 0
                    previous_operator = c
                    
                i[0] += 1
            
            return sum(stack)

        s += "@"
        return solve([0])

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(n)$

Where $n$ is the length of the expression