<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) {
        string dum="";
        vector<string>v;
         int min=strs[0].size();
        for(int i=0;i<strs.size();i++){
            if(min>strs[i].size())
            min=strs[i].size();
        }
            for(int i=0;i<min;i++){
                // for(int j=i;j<min;j++){

                int cnt=0;
               string dum=strs[0].substr(0,i+1);
                for(int k=0;k<strs.size();k++){
                    if(dum==strs[k].substr(0,i+1)){
                        cnt++;
                    
                    }
                }
                if(cnt==strs.size())
                    v.push_back(dum);
                   
               
                // }
            }
        
        int max=1;
        for(int i=0;i<v.size();i++){
            if(max<v[i].size())
            max=v[i].size();
        }
        string ans="";
        for(int i=0;i<v.size();i++){
            if(max==v[i].size()){
            ans=v[i];
            break;
            }
        }
        for(int i=0;i<v.size();i++)
        cout<<v[i]<<"";
        return ans;
    }
};
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