<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>& d) {
        int n = s.size();
        int ans = -1;

        for(int k = 0;k<d.size(); k++){

            string t = d[k];
            int j = 0;

            for(int i = 0; i<n; i++){
                if(s[i] == t[j]){
                    j++;
                }

                if(j == t.size()){
                    break;
                }
            }
            if(j == t.size()){
                if(ans == -1)ans = k;
                else {
                    if(d[ans].size() < t.size()){
                        ans = k;
                    }
                    else if(d[ans].size() == t.size()){
                        if(d[ans] > t){
                            ans = k;
                        }
                    }
                } 
            }
        }
        return ans == -1 ? "" : d[ans];
    }
};
20695