<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 s, vector<string>& dictionary) {
        string longest = "";

        for (const string& word : dictionary) {
            int i = 0, j = 0;

            // Two pointer technique to check if 'word' can be formed from 's'
            while (i < s.size() && j < word.size()) {
                if (s[i] == word[j]) {
                    j++;
                }
                i++;
            }

            // If we've matched all characters of 'word' (i.e., j == word.size())
            if (j == word.size()) {
                // Update the longest word based on length or lexicographical order
                if (word.size() > longest.size() || (word.size() == longest.size() && word < longest)) {
                    longest = word;
                }
            }
        }

        return longest;
    }
};

18299