<p>Given a string <code>s</code> and a string array <code>dictionary</code>, return <em>the longest string in the dictionary that can be formed by deleting some of the given string characters</em>. If there is more than one possible result, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.</p>

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

<pre>
<strong>Input:</strong> s = &quot;abpcplea&quot;, dictionary = [&quot;ale&quot;,&quot;apple&quot;,&quot;monkey&quot;,&quot;plea&quot;]
<strong>Output:</strong> &quot;apple&quot;
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abpcplea&quot;, dictionary = [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;]
<strong>Output:</strong> &quot;a&quot;
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>1 &lt;= dictionary.length &lt;= 1000</code></li>
	<li><code>1 &lt;= dictionary[i].length &lt;= 1000</code></li>
	<li><code>s</code> and <code>dictionary[i]</code> consist of lowercase English letters.</li>
</ul>

class Solution {
public:

    bool isSubsequence(string s, string str){
    int i = 0, j = 0;
    if (str.length() > s.length())
    {
        return false;
    }
    
    while (i < s.length() && j < str.length())
    {
        if (s[i] == str[j])
        {
            j++;
        }
        i++;
    }
    return j==str.length();
}

    string findLongestWord(string s, vector<string>& dictionary) {
        string res = "";
        for (string str : dictionary)
        {
            if ((str.size() > res.size()) || (str.size() == res.size() && str < res))
            {
                if (isSubsequence(s, str))
                {
                    res = str;
                }
            }
            
        }
        return res;
    }
};
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