Leetcode 75

Solution/435. Non-overlapping Intervals/435. Non-overlapping Intervals.py

@@ -0,0 +1,10 @@
+class Solution:
+    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
+        intervals.sort(key=lambda x: x[1])
+        ans = len(intervals)
+        pre = -inf
+        for l, r in intervals:
+            if pre <= l:
+                ans -= 1
+                pre = r
+        return ans

Solution/435. Non-overlapping Intervals/readme.md

@@ -0,0 +1,142 @@
+---
+comments: true
+difficulty: Medium
+edit_url: Antim
+tags:
+    - Greedy
+    - Array
+    - Dynamic Programming
+    - Sorting
+---
+
+<!-- problem:start -->
+
+# [435. Non-overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals)
+
+## Description
+
+<!-- description:start -->
+
+<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>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Sorting + Greedy
+
+We first sort the intervals in ascending order by their right boundary. We use a variable $\textit{pre}$ to record the right boundary of the previous interval and a variable $\textit{ans}$ to record the number of intervals that need to be removed. Initially, $\textit{ans} = \textit{intervals.length}$.
+
+Then we iterate through the intervals. For each interval:
+
+-   If the left boundary of the current interval is greater than or equal to $\textit{pre}$, it means that this interval does not need to be removed. We directly update $\textit{pre}$ to the right boundary of the current interval and decrement $\textit{ans}$ by one;
+-   Otherwise, it means that this interval needs to be removed, and we do not need to update $\textit{pre}$ and $\textit{ans}$.
+
+Finally, we return $\textit{ans}$.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$, where $n$ is the number of intervals.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
+        intervals.sort(key=lambda x: x[1])
+        ans = len(intervals)
+        pre = -inf
+        for l, r in intervals:
+            if pre <= l:
+                ans -= 1
+                pre = r
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public int eraseOverlapIntervals(int[][] intervals) {
+        Arrays.sort(intervals, (a, b) -> a[1] - b[1]);
+        int ans = intervals.length;
+        int pre = Integer.MIN_VALUE;
+        for (var e : intervals) {
+            int l = e[0], r = e[1];
+            if (pre <= l) {
+                --ans;
+                pre = r;
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
+        ranges::sort(intervals, [](const vector<int>& a, const vector<int>& b) {
+            return a[1] < b[1];
+        });
+        int ans = intervals.size();
+        int pre = INT_MIN;
+        for (const auto& e : intervals) {
+            int l = e[0], r = e[1];
+            if (pre <= l) {
+                --ans;
+                pre = r;
+            }
+        }
+        return ans;
+    }
+};
+```
+
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->