Maximize the Number of Partitions After Operations — C++

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

// prefix sum, greedy
class Solution {
public:
    int maxPartitionsAfterOperations(string s, int k) {
        vector<int> left(size(s) + 1);
        vector<int> left_mask(size(s) + 1);
        for (int i = 0, cnt = 0, mask = 0; i < size(s); ++i) {
            mask |= 1 << (s[i] - 'a');
            if (__builtin_popcount(mask) > k) {
                ++cnt;
                mask = 1 << (s[i] - 'a');
            }
            left[i + 1] = cnt;
            left_mask[i + 1] = mask;
        }
        vector<int> right(size(s) + 1);
        vector<int> right_mask(size(s) + 1);
        for (int i = size(s) - 1, cnt = 0, mask = 0; i >= 0; --i) {
            mask |= 1 << (s[i] - 'a');
            if (__builtin_popcount(mask) > k) {
                ++cnt;
                mask = 1 << (s[i] - 'a');
            }
            right[i] = cnt;
            right_mask[i] = mask;
        }

        int result = 0;
        for (int i = 0; i < size(s); ++i) {
            int curr = left[i] + right[i + 1];
            const int mask = left_mask[i] | right_mask[i + 1];
            if (__builtin_popcount(left_mask[i]) == k && __builtin_popcount(right_mask[i + 1]) == k && __builtin_popcount(mask) != 26) {
                curr += 3;
            } else if (__builtin_popcount(mask) + (__builtin_popcount(mask) != 26 ? 1 : 0) > k) {  // test case: s = "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz", k = 26
                curr += 2;
            } else {
                curr += 1;
            }
            result = max(result, curr);
        }
        return result;
    }
};