Add 'Non-overlapping Intervals'

Author: nickolashkraus

README.md

@@ -80,6 +80,8 @@ A collection of LeetCode solutions
 
 [Next Permutation](./src/next_permutation.py)
 
+[Non Overlapping Intervals](./src/non_overlapping_intervals.py)
+
 [Number of Visible People in a Queue](./src/number_of_visible_people_in_a_queue.py)
 
 [Palindromic Substrings](./src/palindromic_substrings.py)

src/non_overlapping_intervals.py

@@ -0,0 +1,28 @@
+"""
+435. Non-overlapping Intervals
+
+https://leetcode.com/problems/non-overlapping-intervals
+
+NOTES
+  * A common routine for interval problems is to sort the array of intervals
+    by each interval's starting index.
+
+We start by sorting the interval array by starting index (O(nlog n)). Since the
+starting index is now monotonically increasing, we remove overlapping intervals
+while taking the lower ending index until there is no overlap.
+"""
+
+import sys
+
+
+class Solution:
+    def eraseOverlapIntervals(self, intervals: list[list[int]]) -> int:
+        intervals.sort(key=lambda x: x[0])
+        count, lo = 0, -sys.maxsize
+        for interval in intervals:
+            if interval[0] < lo:
+                count += 1
+                lo = min(lo, interval[1])
+            else:
+                lo = interval[1]
+        return count

tests/test_non_overlapping_intervals.py

@@ -0,0 +1,23 @@
+"""
+435. Non-overlapping Intervals
+
+https://leetcode.com/problems/non-overlapping-intervals
+"""
+
+from unittest import TestCase
+
+from src.non_overlapping_intervals import Solution
+
+
+class TestSolution(TestCase):
+    def test_1(self):
+        exp = 1
+        assert Solution().eraseOverlapIntervals([[1, 2], [2, 3], [3, 4], [1, 3]]) == exp
+
+    def test_2(self):
+        exp = 2
+        assert Solution().eraseOverlapIntervals([[1, 2], [1, 2], [1, 2]]) == exp
+
+    def test_3(self):
+        exp = 0
+        assert Solution().eraseOverlapIntervals([[1, 2], [2, 3]]) == exp