It is sufficient to implement as instructed in the problem statement. Among the several possible approaches, the following implementation adopts the following one:

• Construct a frequency array (that manages how many times each letter occurs in \(S\)).
• Find the maximum value in the frequency (= maximum frequency).
• Concatenate the characters that do not take the maximum.

Sample code in Python:

s = input()
cnt = {c : s.count(c) for c in s}
ma = max(cnt.values())
print("".join(c for c in s if cnt[c] != ma))

Sample code in C++

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

int main(){
    string s; cin >> s;
    vector<int> cnt(26, 0);
    for (char c : s) cnt[c-'a']++;
    int ma = *max_element(cnt.begin(), cnt.end());
    string t = "";
    for (char c : s){
        if (cnt[c-'a'] != ma) t += c;
    }
    cout << t << endl;
}