<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 memo[50001];
// //     int n;
// //     //find the first job jiska starting point >= currentJob ka end point
// //     int getNextIndex(vector<vector<int>>& array, int l, int currentJobEnd) {
// //         int r = n-1;
        
// //         int result = n+1;
        
// //         while(l <= r) {
// //             int mid = l + (r-l)/2;
            
// //             if(array[mid][0] >= currentJobEnd) { //we can take this task
// //                 result = mid;
// //                 r = mid-1;
// //             } else {
// //                 l = mid+1;
// //             }
// //         }
        
// //         return result;
// //     }
    
// //     int solve(vector<vector<int>>& array, int i) {
// //         if(i >= n)
// //             return 0;
        
// //         if(memo[i] != -1)
// //             return memo[i];
        
// //         int next  = getNextIndex(array, i+1, array[i][1]);
        
// //         int taken = array[i][2] + solve(array, next);
// //         int notTaken = solve(array, i+1);
        
// //         return memo[i] = max(taken, notTaken);
// //     }
    
// //     int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
// //         n = startTime.size();
        
// //         memset(memo, -1, sizeof(memo));
        
// //         vector<vector<int>> array(n, vector<int>(3, 0)); //{s, e, p}
        
// //         for(int i = 0; i<n; i++) {
// //             array[i][0] = startTime[i];
// //             array[i][1] = endTime[i];
// //             array[i][2] = profit[i];
// //         }
        
// //         auto comp = [&](auto& vec1, auto& vec2) {
// //               return vec1[0] <= vec2[0];
// //         };
        
// //         //sort kardo according to sart time
// //         sort(begin(array), end(array), comp);
        
// //         return solve(array, 0);
// //     }
// // };
// class Solution {
// public:
//     int memo[50001];
//     int n;

//     int getNextIndex(vector<vector<int>>& array, int l, int currentJobEnd) {
//         int r = n-1;
//         int result = n+1;

//         while(l <= r) {
//             int mid = l + (r-l)/2;
//             if(array[mid][0] >= currentJobEnd) { 
//                 result = mid;
//                 r = mid-1;
//             } else {
//                 l = mid+1;
//             }
//         }

//         return result;
//     }

//     int solve(vector<vector<int>>& array, int i) {
//         if(i >= n)
//             return 0;

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

//         int next  = getNextIndex(array, i+1, array[i][1]);
//         if (next == n+1) next = n; 

//         int taken = array[i][2] + solve(array, next);
//         int notTaken = solve(array, i+1);

//         return memo[i] = max(taken, notTaken);
//     }

//     int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
//         n = startTime.size();

//         memset(memo, -1, sizeof(memo));

//         vector<vector<int>> array(n, vector<int>(3, 0)); //{s, e, p}

//         for(int i = 0; i<n; i++) {
//             array[i][0] = startTime[i];
//             array[i][1] = endTime[i];
//             array[i][2] = profit[i];
//         }

//         auto comp = [&](auto& vec1, auto& vec2) {
//               return vec1[0] <= vec2[0];
//         };

//         sort(begin(array), end(array), comp);

//         return solve(array, 0);
//     }
// };
class Solution {
public:
    int binarySearch(const vector<vector<int>>& jobs, int end, int left, int right) {
        int result = -1;  // Initialize result to -1 to handle cases where no job is found

        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (jobs[mid][1] <= end) {
                result = mid;
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }

        return result;
    }

    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        int n = startTime.size();
        int dp[n]; 
        vector<vector<int>> jobs(n, vector<int>(3));

        for(int i = 0; i < n; i++) { 
            jobs[i][0] = startTime[i]; 
            jobs[i][1] = endTime[i]; 
            jobs[i][2] = profit[i]; 
        }

        sort(jobs.begin(), jobs.end(), [](const vector<int>& m, const vector<int>& n) {
            return m[1] < n[1];
        });

        dp[0] = jobs[0][2];

        for (int i = 1; i < n; i++) {
            int prev = 0;

            // Use binary search function to find the previous job
            int lastJobIndex = binarySearch(jobs, jobs[i][0], 0, i - 1);
            if (lastJobIndex != -1) {
                prev = dp[lastJobIndex];
            }

            dp[i] = max(prev + jobs[i][2], dp[i-1]);
        }
        
        return dp[n - 1];
    }
};
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