<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>& dict) {
        int n = s.size();
        vector<vector<int>> st(1, vector<int>(26, -1));
        s = "#"+s;
        for (int i = n-1; i>=0; i--) {
            st.insert(st.begin(), st.front());
            st.front()[s[i+1]-'a'] = i+1;
        }

        string res = "";
        for (auto &d: dict) {
            int i = 0;
            bool ok = true;
            for (auto ch: d) {
                i = st[i][ch-'a'];
                if (i == -1) {
                    ok = false;
                    break;
                }
            }
            if (ok) {
                if (d.size() > res.size() || (d.size() == res.size() && d < res))
                    res = d;
            }
        }
        return res;
    }
};

// using state machine ww
20695