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# Basic Calculator LeetCode Solution

## Problem – Basic Calculator LeetCode Solution

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.

## Basic Calculator LeetCode Solution in Java

``````public static int calculate(String s) {
int len = s.length(), sign = 1, result = 0;
Stack<Integer> stack = new Stack<Integer>();
for (int i = 0; i < len; i++) {
if (Character.isDigit(s.charAt(i))) {
int sum = s.charAt(i) - '0';
while (i + 1 < len && Character.isDigit(s.charAt(i + 1))) {
sum = sum * 10 + s.charAt(i + 1) - '0';
i++;
}
result += sum * sign;
} else if (s.charAt(i) == '+')
sign = 1;
else if (s.charAt(i) == '-')
sign = -1;
else if (s.charAt(i) == '(') {
stack.push(result);
stack.push(sign);
result = 0;
sign = 1;
} else if (s.charAt(i) == ')') {
result = result * stack.pop() + stack.pop();
}

}
return result;
}
``````

## Basic Calculator LeetCode Solution in C++

``````int calculate(string s) {
int total = 0;
vector<int> signs(2, 1);
for (int i=0; i<s.size(); i++) {
char c = s[i];
if (c >= '0') {
int number = 0;
while (i < s.size()  &&  s[i] >= '0')
number = 10 * number + s[i++] - '0';
total += signs.back() * number;
signs.pop_back();
i--;
}
else if (c == ')')
signs.pop_back();
else if (c != ' ')
signs.push_back(signs.back() * (c == '-' ? -1 : 1));
}
}
``````

## Basic Calculator LeetCode Solution in Python

``````def calculate(self, s):
total = 0
i, signs = 0, [1, 1]
while i < len(s):
c = s[i]
if c.isdigit():
start = i
while i < len(s) and s[i].isdigit():
i += 1
total += signs.pop() * int(s[start:i])
continue
if c in '+-(':
signs += signs[-1] * (1, -1)[c == '-'],
elif c == ')':
signs.pop()
i += 1
``````
##### Basic Calculator LeetCode Solution Review:

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This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

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