<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 ans = "";
        queue<pair<int, int>> q; //idx in dict, current char we need
        
        for(int i = 0; i < dictionary.size(); i++){
            q.push({i, 0});
        }

        int not_solved;
        int prev;

        for(auto chr : s){
            prev = -1; //no index is lower
            
            while(true){
                auto[dict_idx, char_idx] = q.front();
                if(dict_idx <= prev){
                    break;
                }
                not_solved = true;
                q.pop();
                if(chr == dictionary[dict_idx][char_idx]){
                    char_idx++;

                    if(char_idx == dictionary[dict_idx].length()){
                        if(ans.length() < char_idx || (ans.length() == char_idx && dictionary[dict_idx].compare(ans) < 0) ){
                            ans = dictionary[dict_idx];
                            not_solved = false;
                        }
                    }
                }
                if(not_solved){
                    q.push({dict_idx, char_idx});
                    prev = dict_idx;
                }
            }

        }
        return ans;

    }
};
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