Longest Common Prefix

Problem:
Write a function to find the longest common prefix string amongst an array of strings.

Solution:
class Solution {
public:
    string findPrefix(string a, string b) {
        int size_a = a.size(), size_b = b.size();
        int size = min(size_a, size_b);
        int commonPrefix = 0;
        string sol;
        
        for (int i = 0; i < size; i++) {
            if (a[i] == b[i])
                commonPrefix++;
            else
                break;
        }
        return a.substr(0, commonPrefix);
    }
    string longestCommonPrefix(vector<string>& strs) {
        int size = strs.size();
        
        if (size == 0)
            return "";
        else if (size == 1)
            return strs[0];
        else if (size == 2)
            return findPrefix(strs[0], strs[1]);
        else {
            // Recursive approach
            /*
            vector<string> first, second;
            string firstHalf, secondHalf;
            for (int i = 0; i < size; i++) {
                if ((i % 2) == 0)
                    first.push_back(strs[i]);
                else
                    second.push_back(strs[i]);
            }
            firstHalf = longestCommonPrefix(first);
            secondHalf = longestCommonPrefix(second);
            return findPrefix(firstHalf, secondHalf);
           */
            
            // Iterative approach
            string sol = strs[0];
            for (int i  = 0; i < size; i++) {
                sol = findPrefix(strs[i], sol);
            }
            return sol;
        } 
    }
};

==== Optimized solution ====
// Check vertical 

class Solution {

public:

    string longestCommonPrefix(vector<string> &strs) {

        if (strs.empty()){return "";}

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

         

        for (int i=0;i<strs[0].size();i++){

            for (int j=0;j<strs.size()-1;j++){

                if (i>=strs[j+1].size() || strs[j][i]!=strs[j+1][i]){

                    return strs[j].substr(0,i);

                }

            }

        }

         

        return strs[0];

    }

};