Longest Word in Dictionary through Deleting
<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>
Array; Two Pointers; String; Sorting
impl Solution {
    pub fn find_longest_word(s: String, dictionary: Vec<String>) -> String {
        let mut ans = String::new();
        for t in dictionary {
            let a = ans.len();
            let b = t.len();
            if Self::check(&t, &s) && (a < b || (a == b && t < ans)) {
                ans = t;
            }
        }
        ans
    }

    fn check(s: &str, t: &str) -> bool {
        let (m, n) = (s.len(), t.len());
        let mut i = 0;
        let mut j = 0;
        let s: Vec<char> = s.chars().collect();
        let t: Vec<char> = t.chars().collect();

        while i < m && j < n {
            if s[i] == t[j] {
                i += 1;
            }
            j += 1;
        }
        i == m
    }
}
