How do you solve Non-overlapping Intervals on LeetCode?

Non-overlapping Intervals (LeetCode #435) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting the intervals by their end times allows us to efficiently determine overlaps.

What is the time complexity of Non-overlapping Intervals?

The optimal solution for Non-overlapping Intervals runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(1) since we only use a few extra variables.

What is the brute force solution for Non-overlapping Intervals?

The brute force approach for Non-overlapping Intervals is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach tries every possible combination of intervals to see which ones can be kept without overlapping. This method is simple but inefficient, as it checks all subsets of intervals. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Non-overlapping Intervals?

Non-overlapping Intervals uses the following concepts: Array, Dynamic Programming, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Non-overlapping Intervals?

Always think about sorting as a first step for interval problems. Explain your thought process clearly; it helps the interviewer understand your approach. Practice similar problems to recognize patterns and improve your problem-solving speed.

What are common mistakes when solving Non-overlapping Intervals?

Failing to sort the intervals before processing them. Not correctly checking for overlaps, especially at the boundaries.

#435

Non-overlapping Intervals

Medium

ArrayDynamic ProgrammingGreedySortingGreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

💡

Intuition

Time O(n log n)Space O(1)

The optimal solution uses a greedy approach by sorting the intervals based on their end times. By always selecting the interval that ends the earliest, we can maximize the number of non-overlapping intervals.

⚙️

Algorithm

4 steps

1Step 1: Sort the intervals by their end times.
2Step 2: Initialize a variable to keep track of the end time of the last added interval.
3Step 3: Iterate through the sorted intervals and count how many intervals overlap with the last added interval.
4Step 4: The number of intervals to remove is the total number of intervals minus the count of non-overlapping intervals.

solution.py14 lines

1# Full working Python code
2
3def eraseOverlapIntervals(intervals):
4    if not intervals:
5        return 0
6    intervals.sort(key=lambda x: x[1])
7    count = 0
8    end = intervals[0][1]
9    for i in range(1, len(intervals)):
10        if intervals[i][0] < end:
11            count += 1
12        else:
13            end = intervals[i][1]
14    return count

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(1) since we only use a few extra variables.

1Sorting the intervals by their end times allows us to efficiently determine overlaps.
2Greedy algorithms often provide optimal solutions for interval scheduling problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.