Algorithm Notes
Summary: Non Overlapping Intervals — notes not yet curated.
Time: Estimate via loops/recurrences; common classes: O(1), O(log n), O(n), O(n log n), O(n^2)
Space: Count auxiliary structures and recursion depth.
Tip: See the Big-O Guide for how to derive bounds and compare trade-offs.
Big-O Guide
Source
"""
Non Overlapping Intervals
TODO: Add problem description
"""
from src.interview_workbook.leetcode._registry import register_problem
from src.interview_workbook.leetcode._types import Category, Difficulty
class Solution:
def solve(self, *args):
"""
Non-overlapping Intervals: Minimum to remove to avoid overlap.
Args:
intervals (List[List[int]])
Returns:
int
"""
(intervals,) = args
if not intervals:
return 0
intervals.sort(key=lambda x: x[1])
count = 0
prev_end = intervals[0][1]
for i in range(1, len(intervals)):
if intervals[i][0] < prev_end:
count += 1
else:
prev_end = intervals[i][1]
return count
def demo():
"""TODO: Implement demo function."""
pass
register_problem(
id=435,
slug="non_overlapping_intervals",
title="Non-overlapping Intervals",
category=Category.INTERVALS,
difficulty=Difficulty.MEDIUM,
tags=["array", "dynamic_programming", "greedy", "sorting"],
url="https://leetcode.com/problems/non-overlapping-intervals/",
notes="",
)