<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 sortbysec(pair<int,int> &a,pair<int,int> &b){
        return a.second<b.second;
    }
    int eraseOverlapIntervals(vector<vector<int>>& a) {
        vector<pair<int,int>> nums(a.size());
        int pos=0;
        for(auto i:a){
            nums[pos].first=i[0];
            nums[pos].second=i[1];
            pos++;
        }
        sort(nums.begin(),nums.end(),sortbysec);
        int res=0,curr=0;
        for(int i=1;i<nums.size();i++){
            if(nums[i].first<nums[curr].second)    res++;
            else    curr=i;
        }
        return res;
    }
};
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