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

        // Later to be used for binary search
        vector<int> startTimes(jobs.size());
        for(int i = 0; i < jobs.size(); ++i) {
            startTimes[i] = jobs[i][0];
        }

        unordered_map<int, int> dp;

        return solve(0, jobs, dp, startTimes);
    }

    int solve(int startingIdx, vector<vector<int>> &jobs, unordered_map<int, int> &dp, vector<int> &startTimes) {
        // Base case, we have no more jobs to take.
        if (startingIdx >= jobs.size()) {
            return 0;
        }

        // If we've found a solution for this index before, return it
        if (dp.find(startingIdx) != dp.end()) {
            return dp[startingIdx];
        }

        int ans = 0;
        
        // We can either take the current job, or skip it.
        int skip = solve(startingIdx + 1, jobs, dp, startTimes);

        // while (newJobIdx < jobs.size() && jobs[newJobIdx][0] < jobs[startingIdx][1]) {
        //     newJobIdx++;
        // }

        // To find the next job to take if we're taking current job, we need to find a new job start time where there's no overlap. 
        // Which means new job start time should be >= current job's end time
        // Rather than doing a linear scan, we can sort the startTimes, and just find a next available start time that is >= current endTime
        // Because the jobs are sorted by start time, we can do a binary search for the next available startTime > current
        int newJobIdx = search(startTimes, startingIdx+1, jobs[startingIdx][1]);
        
        int take = jobs[startingIdx][2] + solve(newJobIdx, jobs, dp, startTimes);

        return dp[startingIdx] = max(take, skip);
    }



    // Binary search for a start time that is >= our last ending time
    int search(vector<int> &startTimes, int startingIdx, int endingTime) {
        int left = startingIdx;
        int right = startTimes.size();

        // If we don't find a valid startTime >= endTime, we can just return out of bound as answer.
        // As after returning out of bound, in the recursive call it will be handled by returning 0
        int ans = startTimes.size();
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (startTimes[mid] >= endingTime) {
                ans = mid;
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return ans;
    }
};
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