Maximum Number of Events That Can Be Attended II — C++

// Time:  O(nlogn + n * k)
// Space: O(n * k)

class Solution {
public:
    int maxValue(vector<vector<int>>& events, int k) {
        sort(begin(events), end(events),
             [](const auto& a, const auto& b) {
                 return a[1] < b[1];
             });
        vector<vector<int>> dp(size(events) + 1, vector<int>(k + 1));
        for (int i = 1; i <= size(events); ++i) {
            auto prev_i_m_1 = distance(cbegin(events),
                                       prev(lower_bound(cbegin(events), cend(events), events[i - 1],
                                            [](const auto& a, const auto& b) {
                                                return a[1] < b[0];
                                            })));
            for (int j = 1; j <= k; ++j) {
                dp[i][j] = max(dp[i - 1][j], dp[prev_i_m_1 + 1][j - 1] + events[i - 1][2]);
            }
        }
        return dp[size(events)][k];
    }
};

// Time:  O(nlogn + n * k)
// Space: O(n * k)
class Solution2 {
public:
    int maxValue(vector<vector<int>>& events, int k) {
        sort(begin(events), end(events));
        vector<vector<int>> dp(size(events) + 1, vector<int>(k + 1));
        for (int i = size(events) - 1; i >= 0; --i) {
            auto next_i = distance(cbegin(events),
                                   prev(upper_bound(cbegin(events), cend(events), events[i],
                                        [](const auto& a, const auto& b) {
                                            return a[1] < b[0];
                                        })));
            for (int j = 1; j <= k; ++j) {
                dp[i][j] = max(dp[i + 1][j], dp[next_i + 1][j - 1] + events[i][2]);
            }
        }
        return dp[0][k];
    }
};