<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) {
        unordered_map<char,vector<int>>m;
        string ans="";

        for(int i=0;i<s.size();i++){
            m[s[i]].push_back(i);
        }

        for(int i=0;i<dictionary.size();i++){
            int prev=-1;
            bool f=1;

            for(int j=0;j<dictionary[i].length();j++){
                vector<int>v=m[dictionary[i][j]];
                int ind=upper_bound(v.begin(),v.end(),prev)-v.begin();
                if(ind == v.size()){
                    f=0;
                    break;
                }
                prev=v[ind];
                //cout<<prev<<" ";
            }
            cout<<endl;
            if(f){
                if(ans.length()<dictionary[i].length()){
                    ans=dictionary[i];
                }
                else if(ans.length() == dictionary[i].length()){
                    ans=min(ans,dictionary[i]);
                }
            }
        }

        return ans;
    }
};
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