<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 Solution1 {
public:
    int jobScheduling(vector<int>& st, vector<int>& et, vector<int>& p) {
        vector<vector<int>> vk;
        int n = st.size();
        for(int i=0; i<n; i++) vk.push_back({st[i],et[i],p[i]});
        sort(vk.begin(), vk.end());
        vector<int> dp(n);
        dp[0] = vk[0][2];
        for(int i=1; i<n; i++){
            dp[i] = vk[i][2];
            for(int j=0; j<i; j++){
                if(vk[i][0]>=vk[j][1]) dp[i] = max(dp[i], dp[j] + vk[i][2]);
            }
        }
        return *max_element(dp.begin(), dp.end());
    }
}; //26/35, later tle;

class Solution {
public:
    static bool mcp(vector<int>&a, vector<int>&b){
        if(a[1]==b[1]) return a[0]<b[0];
        return a[1]<b[1];
    }
    int jobScheduling(vector<int>& st, vector<int>& et, vector<int>& p) {
        vector<vector<int>> vk;
        int n = st.size();
        for(int i=0; i<n; i++) vk.push_back({st[i],et[i],p[i]});
        sort(vk.begin(), vk.end(), mcp);
        vector<int> dp(n);
        dp[0] = vk[0][2];
        for(int i=1; i<n; i++){
            dp[i] = dp[i-1]; // choose to not do the job; dp[i] : max profit so far;
            int j = f(vk,0,i,vk[i][0]);
            dp[i] = max(dp[i], vk[i][2] + ( (j>=0)? dp[j] : 0));
        }
        return dp[n-1];
    }
    int f(vector<vector<int>>& vk, int l, int h, int x){
        
        int m;
        while(l<=h){
            m = (l+h)/2;
            if(vk[m][1]<=x) l = m+1;
            else h = m-1;
        }
        return h;
    }
};
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