<p>Write a function to find the longest common prefix string amongst an array of strings.</p>

<p>If there is no common prefix, return an empty string <code>&quot;&quot;</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> strs = [&quot;flower&quot;,&quot;flow&quot;,&quot;flight&quot;]
<strong>Output:</strong> &quot;fl&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> strs = [&quot;dog&quot;,&quot;racecar&quot;,&quot;car&quot;]
<strong>Output:</strong> &quot;&quot;
<strong>Explanation:</strong> There is no common prefix among the input strings.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= strs.length &lt;= 200</code></li>
	<li><code>0 &lt;= strs[i].length &lt;= 200</code></li>
	<li><code>strs[i]</code> consists of only lowercase English letters.</li>
</ul>

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        unordered_set<char> set;
        int k = strs[0].size();
        string current = "";
        string after = "";
        bool whoops = false;
        vector<string> vect;
        string ans = "";

        if(strs.size() ==1){
            return strs[0];
        }

        for(int i = 0; i < strs.size(); i++){
            k = min(k, (int)strs[i].size());
        }
        
        for(int i = 0; i < k; i++){
            for(int j = 0; j < strs.size() - 1; j++){
                current = strs[j].substr(0,i+1);
                after = strs[j + 1].substr(0,i+1);
                if(current != after){
                    whoops = true;
                    break;
                }
                if(current == after && j == strs.size() - 2){
                    vect.push_back(current);
                    ans = current;
                }
            }
            if(whoops){
                break;
            }

        }

        if(vect.size() <= 0){
            return "";
        }

        //return vect[vect.size() - 1];
        return ans;
    }
};
18200