<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:
    void sortBasedOnAnotherVector(vector<int>& keys, vector<int>& values1, vector<int>& values2) {
        vector<std::tuple<int, int, int>> paired;
        for (size_t i = 0; i < keys.size(); ++i) {
            paired.emplace_back(keys[i], values1[i], values2[i]);
        }

        sort(paired.begin(), paired.end(), [](const tuple<int, int, int>& a, const tuple<int, int, int>& b) {
            return get<0>(a) < get<0>(b);
        });

        for (size_t i = 0; i < paired.size(); ++i) {
            keys[i] = get<0>(paired[i]);
            values1[i] = get<1>(paired[i]);
            values2[i] = get<2>(paired[i]);
        }
    }


    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        cin.tie(nullptr);
        cout.tie(nullptr);
        ios_base::sync_with_stdio(false);
        
        int const n = profit.size();
        //int dp[n];
        int size = 1;
        int m = 0;
        int p = 1;
        int ans = 0;
        vector<int> new_endTime(n,0);
        unordered_map <int,int> idx_to_end;
        unordered_map <int,int> end_to_idx;

        for (int i = 0; i<n; i++){new_endTime[i] = endTime[i];}

        sort(new_endTime.begin(),new_endTime.end());

        

        for (int i = 0; i<n; i++){
            auto it = end_to_idx.find(new_endTime[i]);
            if (it == end_to_idx.end()) {
                idx_to_end[size] = new_endTime[i];
                end_to_idx.insert({new_endTime[i], size});
                size++;
            } 
        }

        sortBasedOnAnotherVector(startTime, endTime, profit);
        
        int past_best[size];

      //  fill(dp,dp+n,0);
        fill(past_best, past_best+size,0);
    

        for (int i = 0; i < n; i++){
            if (idx_to_end[p]<=startTime[i]){
                while ((idx_to_end[p]<=startTime[i])){
                    m = max(m,past_best[p]);
                    p++;
                }
                p--;
            }

            
            ans = max(ans,profit[i] + m);
            past_best[end_to_idx[endTime[i]]] = max(profit[i] + m,past_best[end_to_idx[endTime[i]]]);    
        }

        return ans;
    }

};
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