<p>Given a string <code>formula</code> representing a chemical formula, return <em>the count of each atom</em>.</p>

<p>The atomic element always starts with an uppercase character, then zero or more lowercase letters, representing the name.</p>

<p>One or more digits representing that element&#39;s count may follow if the count is greater than <code>1</code>. If the count is <code>1</code>, no digits will follow.</p>

<ul>
	<li>For example, <code>&quot;H2O&quot;</code> and <code>&quot;H2O2&quot;</code> are possible, but <code>&quot;H1O2&quot;</code> is impossible.</li>
</ul>

<p>Two formulas are concatenated together to produce another formula.</p>

<ul>
	<li>For example, <code>&quot;H2O2He3Mg4&quot;</code> is also a formula.</li>
</ul>

<p>A formula placed in parentheses, and a count (optionally added) is also a formula.</p>

<ul>
	<li>For example, <code>&quot;(H2O2)&quot;</code> and <code>&quot;(H2O2)3&quot;</code> are formulas.</li>
</ul>

<p>Return the count of all elements as a string in the following form: the first name (in sorted order), followed by its count (if that count is more than <code>1</code>), followed by the second name (in sorted order), followed by its count (if that count is more than <code>1</code>), and so on.</p>

<p>The test cases are generated so that all the values in the output fit in a <strong>32-bit</strong> integer.</p>

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

<pre>
<strong>Input:</strong> formula = &quot;H2O&quot;
<strong>Output:</strong> &quot;H2O&quot;
<strong>Explanation:</strong> The count of elements are {&#39;H&#39;: 2, &#39;O&#39;: 1}.
</pre>

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

<pre>
<strong>Input:</strong> formula = &quot;Mg(OH)2&quot;
<strong>Output:</strong> &quot;H2MgO2&quot;
<strong>Explanation:</strong> The count of elements are {&#39;H&#39;: 2, &#39;Mg&#39;: 1, &#39;O&#39;: 2}.
</pre>

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

<pre>
<strong>Input:</strong> formula = &quot;K4(ON(SO3)2)2&quot;
<strong>Output:</strong> &quot;K4N2O14S4&quot;
<strong>Explanation:</strong> The count of elements are {&#39;K&#39;: 4, &#39;N&#39;: 2, &#39;O&#39;: 14, &#39;S&#39;: 4}.
</pre>

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

<ul>
	<li><code>1 &lt;= formula.length &lt;= 1000</code></li>
	<li><code>formula</code> consists of English letters, digits, <code>&#39;(&#39;</code>, and <code>&#39;)&#39;</code>.</li>
	<li><code>formula</code> is always valid.</li>
</ul>

class Solution {
public:
    string countOfAtoms(string formula) 
    {
        stack<map<string, int>> stk;
        map<string, int> curr;
        int n = formula.size();
        for (int i = 0; i < n; ) {
            if (formula[i] == '(') 
            {
                stk.push(curr);
                curr.clear();
                i++;
            } 
            else if (formula[i] == ')') 
            {
                int j = i + 1;
                while (j < n && isdigit(formula[j])) j++;
                int multiplicity = j > i + 1 ? stoi(formula.substr(i + 1, j - i - 1)) : 1;
                i = j;
                if (!stk.empty()) 
                {
                    auto top = curr;
                    curr = stk.top();
                    stk.pop();
                    for (auto &p : top) 
                    {
                        curr[p.first] += p.second * multiplicity;
                    }
                }
            } 
            else 
            {
                int j = i + 1;
                while (j < n && islower(formula[j])) j++;
                string atom = formula.substr(i, j - i);
                i = j;
                while (j < n && isdigit(formula[j])) j++;
                int count = j > i ? stoi(formula.substr(i, j - i)) : 1;
                i = j;
                curr[atom] += count;
            }
        }
        while (!stk.empty()) 
        {
            auto top = curr;
            curr = stk.top();
            stk.pop();
            for (auto &p : top) 
            {
                curr[p.first] += p.second;
            }
        }
        string result;
        for (auto it = curr.begin(); it != curr.end(); it++) 
        {
            result += it->first;
            if (it->second > 1) result += to_string(it->second);
        }
        return result;
    }
};
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