[LeetCode Sync] Runtime - 67 ms (84.11%), Memory - 51.4 MB (22.31%)

Author: NinePiece2

Solutions/0435-non-overlapping-intervals/README.md

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+<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><strong>Note</strong> that intervals which only touch at a point are <strong>non-overlapping</strong>. For example, <code>[1, 2]</code> and <code>[2, 3]</code> are non-overlapping.</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>

Solutions/0435-non-overlapping-intervals/solution.py

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+class Solution:
+    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
+        intervals.sort(key=lambda x: x[1])
+        result = len(intervals)
+        last_val = float('-inf')
+
+        for start, end in intervals:
+            if last_val <= start:
+                result -= 1
+                last_val = end
+        
+        return result