<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 s) {
        int n = s.size();
        if(!n)
            return "";
        stack<map<string,int>>st;
        st.push({{}});
        string name;
        int count = 0 ;
        for(int i = 0;i<n;i++){
            char c = s[i];
            if(c>='A' && c<='Z'){
                if(name.size())
                    st.top()[name] += max(1, count);
                name = c;
                count = 0;    
            }
            else if(c>='a' && c<='z')
                name += c;
            else if(isdigit(c))
                count = count*10+c-'0';
            else if(c == '('){
                if(name.size())
                    st.top()[name] += max(1, count);
                name = "";
                count = 0;
                st.push({{}});
            }
            else if(c == ')'){
                if(name.size())
                    st.top()[name] += max(1, count);
                name = "";
                count = 0;
                
                int val = 0;
                while(i+1<n && isdigit(s[i+1]))
                    val = val*10+s[++i]-'0';
                
                auto pre = st.top();
                st.pop();
                merge(st.top(), pre, max(1, val));
            }
        }
        if(name.size())
            st.top()[name] += max(1, count);
        string ret;
        for(auto const&[s, i]:st.top()){
            ret += s+(i>1?to_string(i):"");
        }
        return ret;
    }
    
    void merge(map<string,int>&lhs, map<string,int>&rhs, int count){
        for(const auto&[i, j]:rhs){
            lhs[i] += j*count;
        }
    }
    
};
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