To solve this problem, we need to determine the minimum number of intervals to remove to make the remaining intervals non-overlapping. This problem can be efficiently solved using a greedy algorithm.

Approach

The key insight here is that the maximum number of non-overlapping intervals can be found by selecting intervals that end the earliest. This allows us to leave more room for subsequent intervals. The minimum number of intervals to remove is then the total number of intervals minus the maximum number of non-overlapping intervals.

Steps:

1. Sort Intervals: First, sort the intervals by their end time. This helps in selecting the interval that ends the earliest.
2. Count Non-overlapping Intervals: Iterate through the sorted intervals, selecting each interval if it starts after the end of the last selected interval.
3. Calculate Removals: Subtract the count of non-overlapping intervals from the total number of intervals to get the number of intervals to remove.

Solution Code

def eraseOverlapIntervals(intervals):
    if not intervals:
        return 0

    # Sort intervals by their end time
    intervals.sort(key=lambda x: x[1])

    count_non_overlap = 1
    last_end = intervals[0][1]

    for s, e in intervals[1:]:
        if s >= last_end:
            count_non_overlap += 1
            last_end = e

    return len(intervals) - count_non_overlap

Explanation

1. Sorting: By sorting intervals based on their end time, we ensure that we always consider the interval that frees up space the earliest.
2. Greedy Selection: We start with the first interval (since it ends earliest) and then select each subsequent interval that does not overlap with the last selected interval. This maximizes the number of non-overlapping intervals.
3. Result Calculation: The number of intervals to remove is the total intervals minus the count of non-overlapping intervals, as these are the intervals that need to be removed to make the rest non-overlapping.

This approach runs in O(n log n) time due to sorting, where n is the number of intervals. This is optimal for this problem.

Example: For intervals [[1,2],[2,3],[3,4],[1,3]], the maximum non-overlapping count is 3, so the minimum removals are 4-3=1. The interval [1,3] is the one to remove.

This solution efficiently handles all edge cases, including empty intervals and overlapping intervals. It is a classic application of the greedy algorithm in interval scheduling.

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