<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:

// THIS IS THE BINARY SEARCH APP SO USE TIME COMP ===> O(N * LOG N)

vector<int> dp;

static bool comp(pair<int,pair<int, int>> &a , pair<int,pair<int,int>> &b){
             
             if(a.second.first < b.second.first)
             return true;
                return false;
}

// Find the latest job (in sorted array) that doesn't
// conflict with the job[i]. If there is no compatible job,
// then it returns -1.

int latestNonConflict(vector<pair<int, pair<int, int>>> &arr  , int i)
{
     int curr_start_time = arr[i].first;

    // Custom binary search
    int low = 0, high = i - 1;

    while (low <= high) {
        int mid = (low + high) / 2;

        if (arr[mid].second.first <= curr_start_time) {
            // Check if the next job's finish time is greater than current start time
            if (arr[mid + 1].second.first <= curr_start_time && mid + 1 <= i - 1) {
                low = mid + 1;
            } else {
                return mid;
            }
        } else {
            high = mid - 1;
        }
    }

    return -1;
}

int solve(vector<pair<int, pair<int, int>>> &v, int i){
            
            if(i < 0)return 0;

       if(dp[i] != -1)return dp[i];

        int exclude = solve(v , i - 1);

        int include = v[i].second.second;

       int j = latestNonConflict(v , i);
        if(i != -1)  include += solve(v , j );

        return (dp[i] = max(include , exclude));
}

    int jobScheduling(vector<int>& s, vector<int>& e, vector<int>& p) {

        vector<pair<int, pair<int, int>>> v;
        int n=s.size();

        dp.resize(n,  -1);

        for(int i=0; i<n; i++){
            v.push_back({s[i], {e[i], p[i]}});
        }

        sort(v.begin(), v.end() , comp);
        
        return solve(v, n-1);
    }
};
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