<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 {
public:
   int calculate(string s) {
       stack<pair<int,int>> st; // pair(prev_calc_value , sign before next bracket () )
       
       long long int sum = 0;
       int sign = +1;
       
       for(int i = 0 ; i < s.size() ; ++i)
       {
           char ch = s[i];
           
           if(isdigit(ch))
           {
               long long int num = 0;
               while(i < s.size() and isdigit(s[i]))
               {
                   num = (num * 10) + s[i] - '0';
                   i++;
               }
               i--; // as for loop also increase i , so if we don't decrease i here a sign will be skipped
               sum += (num * sign);
               sign = +1; // reseting sign
           }
           else if(ch == '(')
           {
               // Saving current state of (sum , sign) in stack
               st.push(make_pair(sum , sign));
               
               // Reseting sum and sign for inner bracket calculation
               sum = 0; 
               sign = +1;
           }
           else if(ch == ')')
           {
               sum = st.top().first + (st.top().second * sum);
               st.pop();
           }
           else if(ch == '-')
           {
               // toggle sign
               sign = (-1 * sign);
           }
       }
       return sum;
   }
};
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