## DEV Community

Abhishek Chaudhary

Posted on

# Basic Calculator

Given a string `s` representing a valid expression, implement a basic calculator to evaluate it, and return the result of the evaluation.

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 = " 2-1 + 2 "
Output: 3

Example 3:

Input: s = "(1+(4+5+2)-3)+(6+8)"
Output: 23

Constraints:

• `1 <= s.length <= 3 * 105`
• `s` consists of digits, `'+'`, `'-'`, `'('`, `')'`, and `' '`.
• `s` represents a valid expression.
• `'+'` is not used as a unary operation (i.e., `"+1"` and `"+(2 + 3)"` is invalid).
• `'-'` could be used as a unary operation (i.e., `"-1"` and `"-(2 + 3)"` is valid).
• There will be no two consecutive operators in the input.
• Every number and running calculation will fit in a signed 32-bit integer.

SOLUTION:

``````class Solution:
def eval(self, a, op, b):
if op == '+':
return a + b
elif op == '-':
return a - b
elif op == '/':
return int(a / b)
elif op == '*':
return a * b

def calculate(self, s: str) -> int:
s += " "
precedence = {
'*': 1,
'/': 1,
'+': 0,
'-': 0
}
brackets = {
'(': 1,
')': -1
}
numstack = []
opstack = []
chunk = ""
latestOpenBracket = True
for c in s:
if c == " " or c in precedence or c in brackets:
if len(chunk) > 0:
numstack.append(int(chunk))
chunk = ""
if c in brackets:
if brackets[c] == 1:
opstack.append('(')
latestOpenBracket = True
elif brackets[c] == -1:
latestOpenBracket = False
while opstack[-1] != '(':
currop = opstack.pop()
b = numstack.pop()
a = numstack.pop()
res = self.eval(a, currop, b)
numstack.append(res)
opstack.pop()
elif c in precedence:
if latestOpenBracket:
numstack.append(0)
while len(opstack) > 0 and precedence.get(opstack[-1], -1) >= precedence[c]:
currop = opstack.pop()
b = numstack.pop()
a = numstack.pop()
res = self.eval(a, currop, b)
numstack.append(res)
opstack.append(c)
latestOpenBracket = False
else:
chunk += c
latestOpenBracket = False
while len(opstack) > 0:
currop = opstack.pop()
b = numstack.pop()
a = numstack.pop()
res = self.eval(a, currop, b)
numstack.append(res)
return numstack[0]
``````