<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:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:

        # Idea: sort it, and try to be greedy

        # Observation: 
        # - for the same start, remove the ones that have larger endings
        # - when adding one the the result, if overlap, remove the ones with larger
        #   endings

        # use a heapq
        pq = intervals
        heapq.heapify(pq)

        last = heapq.heappop(pq)
        ans = 0
        while pq:
            interval = heapq.heappop(pq)
            if interval[0] == last[0]: # same start
                if interval[1] < last[1]:
                    last = interval 
                ans += 1
            else: # different start, interval[0] > result[-1][0]
                if interval[0] >= last[1]: # non-overlapping
                    last = interval
                else: # overlapping
                    if interval[1] < last[1]:
                        last = interval 
                        ans += 1
                    else:
                        ans += 1
        return ans
        
45500