<p>Given an array of intervals <code>intervals</code> where <code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code>, return <em>the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping</em>.</p>

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

<pre>
<strong>Input:</strong> intervals = [[1,2],[2,3],[3,4],[1,3]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> [1,3] can be removed and the rest of the intervals are non-overlapping.
</pre>

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

<pre>
<strong>Input:</strong> intervals = [[1,2],[1,2],[1,2]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> You need to remove two [1,2] to make the rest of the intervals non-overlapping.
</pre>

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

<pre>
<strong>Input:</strong> intervals = [[1,2],[2,3]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> You don&#39;t need to remove any of the intervals since they&#39;re already non-overlapping.
</pre>

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

<ul>
	<li><code>1 &lt;= intervals.length &lt;= 10<sup>5</sup></code></li>
	<li><code>intervals[i].length == 2</code></li>
	<li><code>-5 * 10<sup>4</sup> &lt;= start<sub>i</sub> &lt; end<sub>i</sub> &lt;= 5 * 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>>& i) {
       sort(i.begin(), i.end(),[&](vector<int> &i, vector<int> &j) {
        return i[1] < j[1];
       });
       vector<vector<int>>dp;
       for(int j=0; j<i.size(); j++){
         dp.push_back({i[j][0], -1});
         dp.push_back({i[j][1], -1});
       }
       sort(dp.begin(), dp.end()); unordered_map<int, int>m;
       for(int j=0; j<dp.size(); j++){
         if(m.find(dp[j][0])==m.end()){m[dp[j][0]]=j;}
       }
       dp[0][1]=0; int k=0;
       for(int j=1; j<dp.size(); j++){
         dp[j][1]=dp[j-1][1];
         while(k<i.size() && dp[j][0]==i[k][1]){
            int t=dp[m[i[k][0]]][1]+1;
            dp[j][1]=max(t,dp[j][1]);
            k++;
         }
       }
       return i.size()-dp[dp.size()-1][1];
    }
};
168280