<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>

#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("O3", "unroll-loops", "-ffloat-store")
#pragma GCC target("avx,mmx,sse2,sse3,sse4")

// "_": A lambda function to enable faster I/O operations
auto _=[]()noexcept{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);return 0;}();

#include <string>
#include <stack>
#include <any>


using namespace std;

class Solution {
public:
    int calculate(string s) {
        stack<any> stack;
        int num = 0;
        int sign = 1;

        for (char c: s) {
            if (isdigit(c)) {
                num = 10 * num + int(c - '0');
            } else if (c == '+' || c == '-') {
                stack.push(sign * num);
                num = 0;
                sign = (c == '+') ? 1 : -1;
            } else if (c == '(') {
                stack.push(sign);
                stack.push('(');
                sign = 1;
            } else if (c == ')') {
                stack.push(sign * num);
                num = 0;
                while (!stack.empty()) {
                    any val = stack.top();
                    if (val.type() == typeid(char)) {
                        if (any_cast<char>(val) == '(')
                            break;
                    } else if (val.type() == typeid(int)) {
                        num += any_cast<int>(val);
                        stack.pop();
                    }
                }

                if (!stack.empty()) {
                    any val = stack.top();
                    if (val.type() == typeid(char)) {
                        if (any_cast<char>(val) == '(') {
                            stack.pop();
                        }
                    }
                }

                if (!stack.empty()) {
                    sign = any_cast<int>(stack.top());
                    stack.pop();
                }
            }
        }

        int sumOfStack = 0;
        while (!stack.empty()) {
            sumOfStack += any_cast<int>(stack.top());
            stack.pop();
        }
        
        return (sign * num) + sumOfStack;
    }
};
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