<p>You are given an array of <code>events</code> where <code>events[i] = [startDay<sub>i</sub>, endDay<sub>i</sub>, value<sub>i</sub>]</code>. The <code>i<sup>th</sup></code> event starts at <code>startDay<sub>i</sub></code><sub> </sub>and ends at <code>endDay<sub>i</sub></code>, and if you attend this event, you will receive a value of <code>value<sub>i</sub></code>. You are also given an integer <code>k</code> which represents the maximum number of events you can attend.</p>

<p>You can only attend one event at a time. If you choose to attend an event, you must attend the <strong>entire</strong> event. Note that the end day is <strong>inclusive</strong>: that is, you cannot attend two events where one of them starts and the other ends on the same day.</p>

<p>Return <em>the <strong>maximum sum</strong> of values that you can receive by attending events.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2021/01/10/screenshot-2021-01-11-at-60048-pm.png" style="width: 400px; height: 103px;" /></p>

<pre>
<strong>Input:</strong> events = [[1,2,4],[3,4,3],[2,3,1]], k = 2
<strong>Output:</strong> 7
<strong>Explanation: </strong>Choose the green events, 0 and 1 (0-indexed) for a total value of 4 + 3 = 7.</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2021/01/10/screenshot-2021-01-11-at-60150-pm.png" style="width: 400px; height: 103px;" /></p>

<pre>
<strong>Input:</strong> events = [[1,2,4],[3,4,3],[2,3,10]], k = 2
<strong>Output:</strong> 10
<strong>Explanation:</strong> Choose event 2 for a total value of 10.
Notice that you cannot attend any other event as they overlap, and that you do <strong>not</strong> have to attend k events.</pre>

<p><strong class="example">Example 3:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2021/01/10/screenshot-2021-01-11-at-60703-pm.png" style="width: 400px; height: 126px;" /></strong></p>

<pre>
<strong>Input:</strong> events = [[1,1,1],[2,2,2],[3,3,3],[4,4,4]], k = 3
<strong>Output:</strong> 9
<strong>Explanation:</strong> Although the events do not overlap, you can only attend 3 events. Pick the highest valued three.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= events.length</code></li>
	<li><code>1 &lt;= k * events.length &lt;= 10<sup>6</sup></code></li>
	<li><code>1 &lt;= startDay<sub>i</sub> &lt;= endDay<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= value<sub>i</sub> &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution {
public:
    int solve(int index, int left, vector<pair<pair<int,int>,int>>& v, int n, vector<vector<int>>& dp, vector<int>& nextIndex) {
        if (index == n || left == 0) {
            return 0;
        }

        if (dp[index][left] != -1) {
            return dp[index][left];
        }

        // Option 1: Skip the current event
        int nottake = solve(index + 1, left, v, n, dp, nextIndex);

        // Option 2: Take the current event and go to the next non-overlapping event
        int take = v[index].second + solve(nextIndex[index], left - 1, v, n, dp, nextIndex);

        // Return the maximum of both choices
        return dp[index][left] = max(take, nottake);
    }

    int maxValue(vector<vector<int>>& events, int k) {
        // Convert events into the desired format
        vector<pair<pair<int,int>,int>> v;
        for (auto& x : events) {
            v.push_back({{x[0], x[1]}, x[2]});
        }

        // Sort events by start time, and if same, by end time
        sort(v.begin(), v.end());

        int n = events.size();
        vector<vector<int>> dp(n, vector<int>(k + 1, -1));

        // Precompute the next non-overlapping event index for each event
        vector<int> nextIndex(n, n);  // n means no further event is possible
        for (int i = 0; i < n; ++i) {
            int next = upper_bound(v.begin(), v.end(), make_pair(make_pair(v[i].first.second + 1, 0), 0)) - v.begin();
            nextIndex[i] = next;
        }

        // Start solving from the first event with k remaining selections
        return solve(0, k, v, n, dp, nextIndex);
    }
};

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