<p>Given a string <code>s</code> representing a valid expression, implement a basic calculator to evaluate it, and return <em>the result of the evaluation</em>.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;1 + 1&quot;
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot; 2-1 + 2 &quot;
<strong>Output:</strong> 3
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;(1+(4+5+2)-3)+(6+8)&quot;
<strong>Output:</strong> 23
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 3 * 10<sup>5</sup></code></li>
	<li><code>s</code> consists of digits, <code>&#39;+&#39;</code>, <code>&#39;-&#39;</code>, <code>&#39;(&#39;</code>, <code>&#39;)&#39;</code>, and <code>&#39; &#39;</code>.</li>
	<li><code>s</code> represents a valid expression.</li>
	<li><code>&#39;+&#39;</code> is <strong>not</strong> used as a unary operation (i.e., <code>&quot;+1&quot;</code> and <code>&quot;+(2 + 3)&quot;</code> is invalid).</li>
	<li><code>&#39;-&#39;</code> could be used as a unary operation (i.e., <code>&quot;-1&quot;</code> and <code>&quot;-(2 + 3)&quot;</code> is valid).</li>
	<li>There will be no two consecutive operators in the input.</li>
	<li>Every number and running calculation will fit in a signed 32-bit integer.</li>
</ul>

class Solution:
    def calculate(self, s: str) -> int:
        '''
        "(1-(4+5-(3+2)+1)-3)+(6+8)"
         1 + (-4-5+3+2-1) -3 + 6 + 8
        '''
        # 1st iteration - get rid of ( and )
        stack = []
        signsStack = [1]
        minusSigns = 0

        i = 0
        while i < len(s):
            currNum = ""
            while i < len(s) and s[i].isnumeric():
                currNum += s[i]
                i += 1

            if currNum:
                stack.append("-" + currNum if minusSigns % 2 != 0 else currNum)
            if i == len(s):
                break

            if s[i] == "(":
                if stack and stack[-1] == "-":
                    stack[-1] = "+"
                    signsStack.append(-1)
                    minusSigns += 1
                else:
                    signsStack.append(1)
            if s[i] == ")":
                sign = signsStack.pop()
                if sign == -1:
                    minusSigns -= 1
            if s[i] in {"+", "-"}:
                stack.append(s[i])
            i += 1
            
        resp = 0
        op = "+"
        # 2nd iteration through stack and just add or substract numbers
        for x in stack:
            if x in {"+", "-"}:
                op = x
            else:
                num = int(x)
                resp = resp + num if op == "+" else resp - num






        return resp
        


            

        
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