To optimize our solution further, we can use a boolean array to check whether a character has been seen before or not.

Initialize a boolean array of size 26 as false to represent the 26 lowercase alphabets.
Start iterating from the first character. If the character is marked as false, it means it is appearing for the time. Thus, we can add it to our result string and mark it as true.
If a character is already marked true, it means this is a duplicate character. Thus, we can simply skip this character as it is already present in the result string.
Once the iteration is over, return the result string.

Code

#include <bits/stdc++.h>
using namespace std;

// Solution class to remove duplicates using boolean array
class Solution {
public:
    // Function to remove duplicates
    string removeDuplicates(string s) {
        // Create a boolean array to track visited characters
        bool seen[26] = {false};
        string result = "";

        // Traverse the string
        for (char ch : s) {
            // If character is not seen before
            if (!seen[ch - 'a']) {
                // Mark it as seen
                seen[ch - 'a'] = true;
                // Append to result
                result += ch;
            }
        }
        return result;
    }
};

// Main function
int main() {
    string s = "bcabc";
    Solution sol;
    cout << sol.removeDuplicates(s);
    return 0;
}