Non Overlapping Intervals
Detailed guide and Python implementation for the 'Non Overlapping Intervals' problem.
1. Concept Overview
The 'Non Overlapping Intervals' problem is a key challenge in the Intervals section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Non Overlapping Intervals.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Non Overlapping Intervals carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
Write a function eraseOverlapIntervals(intervals: List[List[int]]) -> int.
- •1 <= len(intervals) <= 10^5
- •intervals[i].length == 2
- •-5 * 10^4 <= starti < endi <= 5 * 10^4
Examples
intervals = [[1,2],[2,3],[3,4],[1,3]]
1
[1,3] can be removed and the rest are non-overlapping.
intervals = [[1,2],[1,2],[1,2]]
2
Remove two [1,2] to make the rest non-overlapping.
intervals = [[1,2],[2,3]]
0
Already non-overlapping.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
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Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def erase_overlap_intervals_opt(intervals):
intervals.sort()
res = 0; prevEnd = intervals[0][1]
for start, end in intervals[1:]:
if start >= prevEnd: prevEnd = end
else: res += 1; prevEnd = min(end, prevEnd)
return resBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def erase_overlap_intervals_brute(intervals):
intervals.sort()
def solve(i, prev):
if i == len(intervals): return 0
res = solve(i + 1, prev)
if prev == -1 or intervals[i][0] >= intervals[prev][1]:
res = max(res, 1 + solve(i + 1, i))
return res
return len(intervals) - solve(0, -1)Algorithm Pattern Checklist
When dealing with Intervals data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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