Minimum Operations to Make Character Frequencies Equal — C++

// Time:  O(26 * n)
// Space: O(26)

// freq table, dp
class Solution {
public:
    int makeStringGood(string s) {
        vector<int> cnt(26);
        for (const auto& x : s) {
            ++cnt[x - 'a'];
        }
        int mx = 0, mn = size(s);
        for (const auto& x : cnt) {
            mx = max(mx, x);
            mn = min(mn, x);
        }
        int result = size(s);
        for (int f = mn; f <= mx; ++f) {
            // dp1: min number of the last one of the operations is insert
            // dp2: min number of the last one of the operations is delete
            int dp1 = 0, dp2 = 0;
            for (int i = 0; i < 26; ++i) {
                if (!cnt[i]) {
                    continue;
                }
                int new_dp1 = 0, new_dp2 = 0;
                if (cnt[i] >= f) {
                    new_dp1 = size(s);
                    new_dp2 = min(dp1, dp2) + (cnt[i] - f);
                } else {
                    const int free = i - 1 >= 0 ? (cnt[i - 1] >= f ? cnt[i - 1] - f : cnt[i - 1]) : 0;
                    new_dp1 = min(min(dp1, dp2) + (f - cnt[i]), dp2 + max((f - cnt[i]) - free, 0));
                    new_dp2 = min(dp1, dp2) + cnt[i];
                }
                dp1 = new_dp1;
                dp2 = new_dp2;
            }
            result = min({result, dp1, dp2});
        }
        return result;
    }
};