c++ answer

include

using namespace std;

string ltrim(const string &); string rtrim(const string &);

/* * Complete the 'anagram' function below. * * The function is expected to return an INTEGER. * The function accepts STRING s as parameter. */

int anagram(string s) { int n = s.length();

// If the length is odd, it's impossible to split into two equal halves.
if (n % 2 != 0) {
    return -1;
}

int half_len = n / 2;
string s1 = s.substr(0, half_len);
string s2 = s.substr(half_len);

// Frequency arrays for characters 'a' through 'z'
vector<int> freq1(26, 0);
vector<int> freq2(26, 0);

// Populate frequency for s1
for (char c : s1) {
    freq1[c - 'a']++;
}

// Populate frequency for s2
for (char c : s2) {
    freq2[c - 'a']++;
}

int changes = 0;
// Compare frequencies to find characters that need to be changed in s1
// We only need to count characters that are in s1 but not (or less) in s2
for (int i = 0; i < 26; ++i) {
    if (freq1[i] > freq2[i]) {
        changes += (freq1[i] - freq2[i]);
    }
}

return changes;

}

int main() { ofstream fout(getenv("OUTPUT_PATH"));

string q_temp;
getline(cin, q_temp);

int q = stoi(ltrim(rtrim(q_temp)));

for (int q_itr = 0; q_itr < q; q_itr++) {
    string s;
    getline(cin, s);

    int result = anagram(s);

    fout << result << "\n";
}

fout.close();

return 0;

}

string ltrim(const string &str) { string s(str);

s.erase(
    s.begin(),
    find_if(s.begin(), s.end(), not1(ptr_fun<int, int>(isspace)))
);

return s;

}

string rtrim(const string &str) { string s(str);

s.erase(
    find_if(s.rbegin(), s.rend(), not1(ptr_fun<int, int>(isspace))).base(),
    s.end()
);

return s;

}