<p>We have <code>n</code> jobs, where every job is scheduled to be done from <code>startTime[i]</code> to <code>endTime[i]</code>, obtaining a profit of <code>profit[i]</code>.</p>

<p>You&#39;re given the <code>startTime</code>, <code>endTime</code> and <code>profit</code> arrays, return the maximum profit you can take such that there are no two jobs in the subset with overlapping time range.</p>

<p>If you choose a job that ends at time <code>X</code> you will be able to start another job that starts at time <code>X</code>.</p>

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

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/10/10/sample1_1584.png" style="width: 380px; height: 154px;" /></strong></p>

<pre>
<strong>Input:</strong> startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]
<strong>Output:</strong> 120
<strong>Explanation:</strong> The subset chosen is the first and fourth job. 
Time range [1-3]+[3-6] , we get profit of 120 = 50 + 70.
</pre>

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

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/10/10/sample22_1584.png" style="width: 600px; height: 112px;" /> </strong></p>

<pre>
<strong>Input:</strong> startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]
<strong>Output:</strong> 150
<strong>Explanation:</strong> The subset chosen is the first, fourth and fifth job. 
Profit obtained 150 = 20 + 70 + 60.
</pre>

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

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/10/10/sample3_1584.png" style="width: 400px; height: 112px;" /></strong></p>

<pre>
<strong>Input:</strong> startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]
<strong>Output:</strong> 6
</pre>

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

<ul>
	<li><code>1 &lt;= startTime.length == endTime.length == profit.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= startTime[i] &lt; endTime[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= profit[i] &lt;= 10<sup>4</sup></code></li>
</ul>

/*
方法一：记忆化搜索 + 二分查找

我们先将工作按照开始时间从小到大排序，然后设计一个函数 dfs(i) 表示从第 i 份工作开始，
可以获得的最大报酬。答案即为 dfs(0) 。

函数 dfs(i) 的计算过程如下：

对于第 i 份工作，我们可以选择做，也可以选择不做。如果不做，最大报酬就是 dfs(i+1) ；如果做，
我们可以通过二分查找，找到在第 i 份工作结束时间之后开始的第一份工作，记为 j ，那么最大报酬
就是 profit[i] + dfs(j) 。取两者的较大值即可。即：

dfs(i) = max(dfs(i+1), profit[i] + dfs(j))

其中 j 是满足 startTime[j] ≥ endTime[i] 的最小的下标。

此过程中，我们可以使用记忆化搜索，将每个状态的答案保存下来，避免重复计算。

时间复杂度 O(n × log⁡n) ，其中 n 是工作的数量。
*/
class Solution {
public:
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        int n = profit.size();
        vector<tuple<int, int, int>> jobs(n);
        for (int i = 0; i < n; ++i) jobs[i] = {startTime[i], endTime[i], profit[i]};
        sort(jobs.begin(), jobs.end());
        vector<int> f(n);
        function<int(int)> dfs = [&](int i) -> int {
            if (i >= n) return 0;
            if (f[i]) return f[i];
            auto [_, e, p] = jobs[i];
            tuple<int, int, int> t{e, 0, 0};
            int j = lower_bound(jobs.begin() + i + 1, jobs.end(), t, [&](auto& l, auto& r) -> bool { return get<0>(l) < get<0>(r); }) - jobs.begin();
            int ans = max(dfs(i + 1), p + dfs(j));
            f[i] = ans;
            return ans;
        };
        return dfs(0);
    }
};
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