<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 findPrefixTwo(string a, string b){
    //using two pointer each on a and b
    int i =0; 
    int j = 0;

    string ans = "";

    while(i<a.size() || j<b.size()){
        if(a[i] == b[j]){
            ans = ans+a[i];
            i++, j++;
        }

        else{
            return ans;
        }
    }

    return ans;
}

    string longestCommonPrefix(vector<string>& arr) {
        if(arr.size() == 1){
            return arr[0];
        }

        //compare first two strings
        string prefix2 = findPrefixTwo(arr[0], arr[1]);

        if(arr.size()==2){
            return prefix2;
        }

        //now compare prefixes for third element to last element and update it
        string ans = "";
        for(int i =2; i<arr.size(); i++){
            ans = findPrefixTwo(prefix2, arr[i]);

            prefix2 = ans;
        }
        return ans;
    }
};
17600