<p>There are <code>n</code> different online courses numbered from <code>1</code> to <code>n</code>. You are given an array <code>courses</code> where <code>courses[i] = [duration<sub>i</sub>, lastDay<sub>i</sub>]</code> indicate that the <code>i<sup>th</sup></code> course should be taken <b>continuously</b> for <code>duration<sub>i</sub></code> days and must be finished before or on <code>lastDay<sub>i</sub></code>.</p>

<p>You will start on the <code>1<sup>st</sup></code> day and you cannot take two or more courses simultaneously.</p>

<p>Return <em>the maximum number of courses that you can take</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> courses = [[100,200],[200,1300],[1000,1250],[2000,3200]]
<strong>Output:</strong> 3
Explanation: 
There are totally 4 courses, but you can take 3 courses at most:
First, take the 1<sup>st</sup> course, it costs 100 days so you will finish it on the 100<sup>th</sup> day, and ready to take the next course on the 101<sup>st</sup> day.
Second, take the 3<sup>rd</sup> course, it costs 1000 days so you will finish it on the 1100<sup>th</sup> day, and ready to take the next course on the 1101<sup>st</sup> day. 
Third, take the 2<sup>nd</sup> course, it costs 200 days so you will finish it on the 1300<sup>th</sup> day. 
The 4<sup>th</sup> course cannot be taken now, since you will finish it on the 3300<sup>th</sup> day, which exceeds the closed date.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> courses = [[1,2]]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> courses = [[3,2],[4,3]]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= courses.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= duration<sub>i</sub>, lastDay<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
</ul>

//VERY DIFFICULT

//Mistake in first attempt
//First of all I sorted the vector in increasing durationa and decreasing last day
//But here it is possible that later on courses would have very low lastday with initial courses would have a higher
//one.So here it is not advisable to take the first values before the later ones


//Right Approach
//Always sort according to time.It mainly works.Sort w.r.t last day.
//Noq use priority_queue.Put the lements one by one.
//If the limits exceed,try popping out the maximum element
//else continue
//Ans is pq.size()
//https://www.youtube.com/watch?v=ey8FxYsFAMU

#define ll long long
#define pb push_back
class Solution {
public:
    int scheduleCourse(vector<vector<int>>& c) {
        ll n = c.size();
        vector<vector<ll>>store(n);
        for(ll i=0;i<n;i++){
            store[i]={c[i][1],c[i][0]};
        }
        sort(store.begin(),store.end());
        priority_queue<ll>pq;
        ll maxday = 0;
        for(ll i=0;i<n;i++){
           
            if(maxday+store[i][1]<=store[i][0]){
                pq.push(store[i][1]);
                maxday+=store[i][1];
            }else{
                if(!pq.empty()){
                    cout<<pq.top()<<endl;
                    if(pq.top()>store[i][1] && maxday-pq.top()+store[i][1]<=store[i][0]){
                         maxday=maxday-pq.top()+store[i][1];
                        pq.pop();
                        pq.push(store[i][1]);
                       
                    }
                }
            }
        }
        return pq.size();
    }
};
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