<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 func(vector<vector<int>>& intervals, int ind, int prev_end)
    {
        if(ind == intervals.size())
            return 0;
        // int self = 1e9,other=1e9;
        // if(dp[ind]!=-1)
        //     return dp[ind];
        if(intervals[ind][0]<prev_end)
        {
            return func(intervals,ind+1,prev_end)+1;
        }
        else
        {
            // ints.push_back(intervals[ind]);
            return func(intervals,ind+1,intervals[ind][1]);
            // ints.pop_back();
        }
    }
    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
        if(intervals.size()<2)
            return 0;
        // vector<int> dp(intervals.size(),-1);
        sort(intervals.begin(),intervals.end(),[](const auto& a, const auto& b){
                return a[1]<b[1];
        });
        // vector<vector<int>> ints;
        // ints.push_back(intervals[0]);
        return func(intervals,0,-1e9);
    }
};
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