<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>

class Solution {
public:
    using int3 = array<int, 3>;
    vector<vector<int>> jobs; // (startTime, endTime, profit)
    vector<int> dp;
    int n;
    vector<int> next;

   void binary_search() {
        for (int i = 0; i < n; i++) {
            // Be careful
            next[i] = lower_bound(jobs.begin(), jobs.end(),
                vector<int>{jobs[i][1],0,0}) - jobs.begin();
        //    cout << i << "->" << next[i] << endl;
        }
    }

    int dfs(int i) {
        if (i == n ) return 0;
        if (dp[i]!= -1) return dp[i];

        // take the job i
        int profit_w_i = jobs[i][2] + dfs(next[i]);

        // Skip the job i
        int profit_n_i = dfs(i+1);

        // Choose the maximum of profit_w_i & profit_n_i
        return dp[i] = max(profit_w_i, profit_n_i);
    }

    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) 
    {
        n = startTime.size();
        jobs.resize(n);
        for (int i = 0; i < n; i++)
            jobs[i] = {startTime[i], endTime[i], profit[i]};

        sort(jobs.begin(), jobs.end());
        // Initialize the 'next' vector with the correct size
        next.resize(n, -1);

        dp.resize(n+1, -1);
        binary_search();
        return dfs(0);
    }
};


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