Leetcode #435: Non-overlapping Intervals
In this guide, we solve Leetcode #435 Non-overlapping Intervals in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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. Note that intervals which only touch at a point are non-overlapping.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Array, Dynamic Programming, Sorting
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: intervals = [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.
Python Solution
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
Complexity
The time complexity is , and the space complexity is , where is the number of intervals. The space complexity is , where is the number of intervals.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.