Here is solution in python, java, c++, c and javascript - https://programmingoneonone.com/hackerrank-anagram-problem-solution.html

int anagram(string s) {

if(s.size() % 2 != 0) return -1;

string left = s.substr(0 , s.size()/2);
string right = s.substr(s.size()/2);

int freq[26] = {0};

for(char c : left) freq[c - 'a']++;
for(char c : right) freq[c - 'a']--;

int changes = 0;

for(int i = 0; i < 26 ; i++){
    if(freq[i] > 0) changes += freq[i];
}
 return changes;

}

include

include

using namespace std;

/*Think of it this way: e.g.xaxb|bbxx left : 2x 1a 1b right: 2x 0a 2b , and they cancle out. What is left in the right is the answer.

O(n) per query*/

int arr[26];

int main() { ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);

int q;
cin >> q; 

string str; 
while(q--) {
    cin >> str; 
    if(str.size() % 2) { cout << -1 << '\n'; continue; }

    int mid = str.size() >> 1; 

    for(int i = 0; i < mid; ++i) ++arr[str[i] - 'a']; 

    int ret = 0; 
    for(int i = mid; i < str.size(); ++i) {
        if(arr[str[i] - 'a']) --arr[str[i] - 'a']; 
        else ++ret; 
    }

    cout << ret << '\n'; 
    memset(arr, 0, sizeof(arr)); 
}

return 0; 

}

Minimalist O(n) solution:

    public static int anagram(String s) {
        if (s.length() % 2 != 0) return -1;
        
        int size = s.length() / 2;
        int[] s1 = new int[26];
        int base = 'a';
        int changes = 0;        

        //build first half
        for (int i = 0; i < size; i++) {
            s1[s.charAt(i) - base] ++;
        }

        //check second half
        for (int i = size; i < size*2; i++) {
            if (s1[s.charAt(i) - base] > 0) s1[s.charAt(i) - base]--;
            else changes++;
        }

        return changes;
    
    }