<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:
    static bool cmp(string a,string b)
    {
        if(a.length()>b.length())
            return true;
        else if(a.length()<b.length())
            return false;
        else
        {
            for(int i=0;i<a.length();i++)
            {
                if(a[i]==b[i])
                    continue;
                return a[i]<b[i];
            }
            return true;
        }
    }
    string findLongestWord(string s, vector<string>& dictionary) {
        sort(dictionary.begin(),dictionary.end(),cmp);
        for(int i=0;i<dictionary.size();i++)
        {
            int j=0;
            int k=0;
            while(j<dictionary[i].length() && k<s.length())
            {
                if(s[k]==dictionary[i][j])
                    j++;
                k++;
            }
            if(j==dictionary[i].length())
                return dictionary[i];
        }
        return "";
    }
};
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