<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:
	string findLongestWord(string str, vector<string>& dictionary) 
	{
		vector<vector<int>>charpos(26); 

		for (int i = 0; i < str.size(); i++) {
			char ch = str[i]; 
			charpos[ch - 'a'].push_back(i); 
		}

		string longest(""); 

		for (auto& word : dictionary) {
			int indx = 0; 
			int indxword = 0; 
			while (indxword != -1 && indxword < word.size()) {
				char ch = word[indxword++]; 
				auto list = charpos[ch - 'a']; 
				auto it = lower_bound(list.begin(), list.end(), indx); 
				if (it == list.end())
					indxword = -1; 
				else 
					indx = *it + 1; 
			}
			if (indxword == -1)
				continue; 
			if (word.size() > longest.size() || (word.size() == longest.size() && word < longest))
				longest = word; 
		}

		return longest; 
	}
};
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