<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:
     static bool cmp(vector<int>&a,vector<int>&b){
        return a[0]<b[0];
    }
    int eraseOverlapIntervals(vector<vector<int>>& inter) {
         sort(inter.begin(),inter.end(),cmp);
        vector<vector<int>>ans;
        int n=inter.size();
        ans.push_back(inter[0]);
        int num=0;
        for(int i=1;i<n;i++){
            vector<int>curr=inter[i];
            vector<int>v=ans[ans.size()-1];
            cout<<curr[0]<<" "<<v[1]<<endl;
            if(curr[0]<v[1]) { //merge
               num++;
               vector<int>v2(2,0);
               v2[0]=max(curr[0],v[0]);
               v2[1]=min(curr[1],v[1]);
               ans.push_back(v2);
            }
            else {
                
                ans.push_back(curr);
            }
        }
        
        return num;
    }
};
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