<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:
    // Helper function to compare jobs by their end times
    static bool compare(const pair<int, pair<int, int>>& a, const pair<int, pair<int, int>>& b) {
        return a.first < b.first;
    }
    
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        int n = startTime.size();
        vector<pair<int, pair<int, int>>> jobs;
        
        // Combine startTime, endTime, and profit into a single list of jobs
        for (int i = 0; i < n; i++) {
            jobs.push_back({endTime[i], {startTime[i], profit[i]}});
        }
        
        // Sort jobs by end time
        sort(jobs.begin(), jobs.end(), compare);
        
        // DP array to store maximum profit until the current job
        vector<int> dp(n, 0);
        dp[0] = jobs[0].second.second; // Profit of the first job
        
        // Binary search helper to find the latest job that doesn't overlap
        auto findLastNonConflictingJob = [&](int index) {
            int low = 0, high = index - 1;
            while (low <= high) {
                int mid = low + (high - low) / 2;
                if (jobs[mid].first <= jobs[index].second.first) {
                    if (jobs[mid + 1].first <= jobs[index].second.first) {
                        low = mid + 1;
                    } else {
                        return mid;
                    }
                } else {
                    high = mid - 1;
                }
            }
            return -1;
        };
        
        // Fill the DP array
        for (int i = 1; i < n; i++) {
            // Option 1: Exclude the current job
            int includeProfit = jobs[i].second.second;
            int lastNonConflictingJob = findLastNonConflictingJob(i);
            if (lastNonConflictingJob != -1) {
                includeProfit += dp[lastNonConflictingJob];
            }
            
            // Option 2: Include the current job
            dp[i] = max(dp[i - 1], includeProfit);
        }
        
        // The answer is the maximum profit we can achieve considering all jobs
        return dp[n - 1];
    }
};

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