Introduction

In this blog post, we will explore how to solve the LeetCode problem "1288" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an array intervals where intervals[i] = [li, ri] represent the interval [li, ri), remove all intervals that are covered by another interval in the list. The interval [a, b) is covered by the interval [c, d) if and only if c <= a and b <= d. Return the number of remaining intervals. Example 1: Input: intervals = [[1,4],[3,6],[2,8]] Output: 2 Explanation: Interval [3,6] is covered by [2,8], therefore it is removed. Example 2: Input: intervals = [[1,4],[2,3]] Output: 1 Constraints: 1 <= intervals.length <= 1000 intervals[i].length == 2 0 <= li < ri <= 105 All the given intervals are unique.

Explanation

Here's the solution:

• Sort Intervals: Sort the intervals based on their start times in ascending order. If start times are equal, sort by descending order of end times. This ensures that if one interval covers another, the covering interval will appear earlier in the sorted list.

• Iterate and Check for Coverage: Iterate through the sorted intervals, keeping track of the end time of the previously considered non-covered interval. If the current interval's end time is less than or equal to the previous end time, it's covered and should be skipped. Otherwise, it's not covered, so increment the count of non-covered intervals and update the previous end time.

• Runtime & Storage Complexity: O(n log n) time, O(1) extra space (excluding space for sorting).

Code

    def removeCoveredIntervals(intervals):
    """
    Removes covered intervals from a list of intervals.

    Args:
        intervals: A list of intervals, where each interval is a list [li, ri].

    Returns:
        The number of remaining intervals after removing covered intervals.
    """

    # Sort the intervals based on start time (ascending) and end time (descending)
    intervals.sort(key=lambda x: (x[0], -x[1]))

    count = 0
    prev_end = -1  # Initialize with a value smaller than any possible end

    for interval in intervals:
        start, end = interval
        if end > prev_end:
            count += 1
            prev_end = end

    return count