<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) {
        int n = a + b;

        int numberB;
        char bigger;
        int numberS;
        char smaller;
        
        if(a>b){
            numberB = a;
            bigger = 'a';
            numberS = b;
            smaller = 'b';
        }
        else{
            numberB = b;
            bigger = 'b';
            numberS = a;
            smaller = 'a';
        }

        vector<vector<char>> temp;

        int i=0;

        while(i<numberB-1){
            vector<char> flag = {bigger, bigger};
            temp.push_back(flag);
            i += 2;
        }

        int s = numberB/2;
        if(i==numberB-1){
            vector<char> flag = {bigger};
            temp.push_back(flag);
            s++;
        }

        string ans;

        int two_per_row = (numberS/s == 0) ? 0 : numberS-s;

        int flag = 1;

        for(int i=0; i<s; i++){
            if(i==two_per_row) flag=0;
            ans += temp[i][0];
            if(temp[i].size()==2) ans += temp[i][1];
            if(flag==1 && numberS>0){
                ans += smaller;
                ans += smaller;
                numberS -= 2;
            }
            else if(flag==0 && numberS>0){
                ans += smaller;
                numberS--;
            }
        }

        return ans;
    }
};
8200