<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>> intervals;
        unordered_map<int, int> mp;
        vector<int> keys;

        mp[-1] = 0;

        for (int i = 0; i < endTime.size(); ++i){
            intervals.push_back({startTime[i], endTime[i], profit[i]});
        }

        sort(intervals.rbegin(), intervals.rend());


        int greatestSoFar = 0;
        for (auto i: intervals){
            if (!mp.contains(i[0])) keys.push_back(i[0]);
            int key = BinarySearchGreater(keys, i[1]);
            mp[i[0]] = max(mp[i[0]], max(mp[key] + i[2], greatestSoFar));
            greatestSoFar = max(greatestSoFar, mp[i[0]]);
        }

        int start =  BinarySearchGreater(keys, 0);
        return greatestSoFar;
    }
    int BinarySearchGreater(vector<int> &nums, int key){
        int left = 0;
        int right = nums.size() - 1;
        while (left <= right){
            int mid = left + (right - left)/2;
            if (nums[mid] == key) return key;
            else if (nums[mid] < key) right = mid - 1;
            else left = mid + 1;
        }
        return right >= nums.size()? -1: nums[right];
    }
};
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