string s;
cin >> s;
sort(s.begin(), s.end());
s.erase(unique(s.begin(), s.end()), s.end());
if (s.size() % 2 == 0) {
    cout << "CHAT WITH HER!\n";
} else {
    cout << "IGNORE HIM!\n";
}

Sort the vector (call this vector v)

Then use v.erase(unique(v.begin(), v.end()), v.end());

Then your answer is just the size of the vector afterwards.

Read about erase and unique to understand what they do.

Alternatively you could insert all the elements in a set and return it's size

If you are interested only in the count of unique characters of a string x, it suffices to do the following:

sort(x.begin(), x.end());
int unique_chars_cnt = unique(x.begin(), x.end()) - x.begin();

Anyway I think using a bucket is better than sorting and unique() since we only have lowercase Latin letters.

string s;
cin >> s;
bool bucket[26];
memset(bucket, 0, sizeof(bucket));
int count = 0;
for (char c : s) {
    if (!bucket[c-'a']) {
        bucket[c-'a'] = true;
        count++;
    }
}
cout << (count%2?"IGNORE HIM!":"CHAT WITH HER!") << endl;

One-liner if you're really lazy: set(s.begin(), s.end()).size()