<p>Given two integers <code>a</code> and <code>b</code>, return <strong>any</strong> string <code>s</code> such that:</p>

<ul>
	<li><code>s</code> has length <code>a + b</code> and contains exactly <code>a</code> <code>&#39;a&#39;</code> letters, and exactly <code>b</code> <code>&#39;b&#39;</code> letters,</li>
	<li>The substring <code>&#39;aaa&#39;</code> does not occur in <code>s</code>, and</li>
	<li>The substring <code>&#39;bbb&#39;</code> does not occur in <code>s</code>.</li>
</ul>

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

<pre>
<strong>Input:</strong> a = 1, b = 2
<strong>Output:</strong> &quot;abb&quot;
<strong>Explanation:</strong> &quot;abb&quot;, &quot;bab&quot; and &quot;bba&quot; are all correct answers.
</pre>

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

<pre>
<strong>Input:</strong> a = 4, b = 1
<strong>Output:</strong> &quot;aabaa&quot;
</pre>

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

<ul>
	<li><code>0 &lt;= a, b &lt;= 100</code></li>
	<li>It is guaranteed such an <code>s</code> exists for the given <code>a</code> and <code>b</code>.</li>
</ul>

class Solution {
public:
    string strWithout3a3b(int a, int b) {
        priority_queue<pair<int, char>> pq;
        string ans = "";
        pq.push({a, 'a'});
        pq.push({b, 'b'});

        while(pq.size()>1){
            pair<int, char> pr1=pq.top();
            pq.pop();
            pair<int, char> pr2=pq.top();
            pq.pop();

            if(ans.empty()){
                pr1.first--;
                ans+=pr1.second;
                if(pr1.first>0){
                    pr1.first--;
                    ans+=pr1.second;
                }
            }else{
                if(ans.back() == pr1.second){
                    pr2.first--;
                    ans+=pr2.second;
                }else{
                    pr1.first--;
                    ans+=pr1.second;
                    if(pr1.first>0){
                        pr1.first--;
                        ans+=pr1.second;
                    }
                }
            }

            if(pr1.first>0) pq.push(pr1);
            if(pr2.first>0) pq.push(pr2);
        }
        for(int i=0; i<pq.top().first; i++){
            ans+=pq.top().second;
        }
        return ans;
    }
};
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