<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>

#define vi vector<int>
#define vvi vector<vi>

int helper(vvi &temp, vi &dp, int pos) {
    if (pos>=dp.size()) {
        return 0;
    }
    if (dp[pos] != -1) {
        return dp[pos];
    }
    int endTime = temp[pos][1];
    int ret = temp[pos][2];
    int i = pos;
    while (i<dp.size() && temp[i][0] < endTime) {
        i++;
    }
    // cout<<pos<<" ";
    helper(temp, dp, i);
    ret += i<dp.size() ? (*max_element(dp.begin()+i, dp.end())) : 0;
    return dp[pos] = ret;
}

class Solution {
public:
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        vvi temp;
        for (int i=0; i<startTime.size(); i++) {
            temp.push_back({startTime[i], endTime[i], profit[i]});
        }
        sort(temp.begin(), temp.end());
        vi dp(temp.size(), 0);
        dp.back() = temp.back()[2];
        vi start(temp.size(), 0);
        start.back() = temp.back()[0];
        for (int i=dp.size()-2; i>=0; i--) {
            dp[i] = temp[i][2];
            start[i] = temp[i][0];
            int pos = lower_bound(start.begin(), start.end(), temp[i][1]) - start.begin();
            if (pos < temp.size()) {
                dp[i] += dp[pos];
            }
            if (dp[i+1] > dp[i]) {
                dp[i] = dp[i+1];
                start[i] = start[i+1];
            }
        }
        // for (int i=0; i<dp.size(); i++) {
        //     if (dp[i] == -1) {
        //         dp[i] = helper(temp, dp, i);
        //     }
        //     cout<<dp[i]<<" ";
        // }
        return dp[0];
    }
};
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