How do you solve Remove Covered Intervals on LeetCode?

Remove Covered Intervals (LeetCode #1288) 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 helps in efficiently determining coverage.

What is the time complexity of Remove Covered Intervals?

The optimal solution for Remove Covered Intervals runs in O(n log n) time and uses O(1) space. The complexity is dominated by the sorting step, which is O(n log n). The subsequent iteration is O(n).

What is the brute force solution for Remove Covered Intervals?

The brute force approach for Remove Covered Intervals is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each interval against all others to see if it is covered. This is straightforward but inefficient, as it requires comparing every interval with every other interval. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Remove Covered Intervals?

Remove Covered Intervals uses the following concepts: Array, Sorting. The recommended patterns to study are: Sorting, Greedy Algorithm.

What are the interview tips for Remove Covered Intervals?

Explain your thought process clearly during the interview. Consider edge cases, such as overlapping intervals. Practice sorting and iterating through arrays to improve speed.

What are common mistakes when solving Remove Covered Intervals?

Not sorting the intervals before checking coverage. Confusing the conditions for coverage.

#1288

Remove Covered Intervals

Medium

ArraySortingSortingGreedy Algorithm

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 approach involves sorting the intervals first and then iterating through them to count non-covered intervals. By sorting, we can efficiently determine coverage based on the start and end points.

⚙️

Algorithm

3 steps

1Step 1: Sort the intervals by starting point, and by ending point in descending order if starting points are the same.
2Step 2: Initialize a variable to track the end of the last added interval.
3Step 3: Iterate through the sorted intervals, and for each interval, check if its end is greater than the last added end. If so, increment the count and update the last added end.

solution.py9 lines

1def removeCoveredIntervals(intervals):
2    intervals.sort(key=lambda x: (x[0], -x[1]))
3    count = 0
4    last_end = 0
5    for interval in intervals:
6        if interval[1] > last_end:
7            count += 1
8            last_end = interval[1]
9    return count

ℹ

Complexity note: The complexity is dominated by the sorting step, which is O(n log n). The subsequent iteration is O(n).

1Sorting helps in efficiently determining coverage.
2Using a single pass after sorting reduces complexity.

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