<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 calc(int idx, int k, vector<int> &nxtIdx, vector<vector<int>> &events, vector<vector<int>> &dp) {
        if (idx == events.size() || k <= 0) return 0; 
        if (dp[idx][k] != -1) return dp[idx][k]; 

        int notTake = calc(idx + 1, k, nxtIdx, events, dp); 
        int take = events[idx][2] + calc(nxtIdx[idx], k - 1, nxtIdx, events, dp); 
        return dp[idx][k] = max(notTake, take); 
    }

    int maxValue(vector<vector<int>>& events, int k) {
        int n = events.size(); 

        sort(events.begin(), events.end()); 
        vector<int> nxtIdx(n); 
        for (int i = 0 ; i < n ; i++) {
            int end = events[i][1]; 
            int l = i + 1, r = n; 
            while (l < r) {
                int mid = (l + r) >> 1; 
                if (events[mid][0] > end) {
                    r = mid; 
                } else {
                    l = mid + 1; 
                }
            }
            nxtIdx[i] = r; 
        }

        // vector<vector<int>> dp(n + 1, vector<int>(k + 1, -1)); 
        // return calc(0, k, nxtIdx, events, dp); 

        vector<vector<int>> dp(n + 1, vector<int>(k + 1)); 
        for (int i = n - 1 ; i >= 0 ; i--) {
            for (int j = 1 ; j <= k ; j++) {
                int notTake = dp[i + 1][j]; 
                int take = events[i][2] + dp[nxtIdx[i]][j - 1]; 
                dp[i][j] = max(take, notTake); 
            }
        }

        return dp[0][k]; 
    }
};
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