We use a stack to process the expression. We maintain the current number and the last operator seen. When we encounter a new operator (or the end of the string), we perform the "last operator" operation on the current number and push the result to the stack.

1. num = 0, last_op = '+', stack = [].
2. Iterate through string s:
   • If char is digit, update num.
   • If char is operator or end of string:
     • If last_op == '+': stack.append(num).
     • If last_op == '-': stack.append(-num).
     • If last_op == '*': stack.append(stack.pop() * num).
     • If last_op == '/': stack.append(int(stack.pop() / num)).
     • Update last_op = char, num = 0.
3. Return sum(stack).

• Time Complexity: O(N).
• Space Complexity: O(N).

def calculate(s):
    if not s: return 0
    stack, num, sign = [], 0, "+"
    s += "+" # dummy to trigger last op
    for char in s:
        if char == " ": continue
        if char.isdigit():
            num = num * 10 + int(char)
        else:
            if sign == "+":
                stack.append(num)
            elif sign == "-":
                stack.append(-num)
            elif sign == "*":
                stack.append(stack.pop() * num)
            elif sign == "/":
                top = stack.pop()
                stack.append(int(top / num))
            sign = char
            num = 0
    return sum(stack)