To solve the problem of finding the minimum number of intervals to remove to make all remaining intervals non-overlapping, we can use a greedy algorithm. Here's the step-by-step approach and solution:

Problem Analysis

The key insight is to select the maximum number of non-overlapping intervals. The minimum number of intervals to remove is the total intervals minus this maximum count.

The greedy choice here is to always pick the interval that ends the earliest. This leaves the maximum room for subsequent intervals, allowing us to fit more non-overlapping intervals.

Algorithm Steps

1. Sort Intervals: Sort the intervals by their end time in ascending order.
2. Select Non-overlapping Intervals:
   • Start with the first interval (earliest end time).
   • For each subsequent interval, if it starts after or at the end of the last selected interval, include it and update the last end time.
3. Calculate Removals: Subtract the number of non-overlapping intervals from the total to get the minimum removals.

Code Implementation

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

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

    count = 1  # At least one interval can be kept
    last_end = intervals[0][1]

    for start, end in intervals[1:]:
        if start >= last_end:
            count += 1
            last_end = end

    return len(intervals) - count

Explanation

• Sorting: By sorting intervals by end time, we prioritize intervals that free up space early.
• Greedy Selection: Each time we pick an interval that doesn't overlap with the last selected one, maximizing the number of non-overlapping intervals.
• Result: The difference between total intervals and the maximum non-overlapping count gives the minimum intervals to remove.

Example:
For intervals [[1,4],[2,3],[3,4]], sorted by end: [[2,3], [3,4], [1,4]]. The maximum non-overlapping count is 2, so the minimum removals are 3-2=1.

This approach runs in O(n log n) time (due to sorting), which is optimal for this problem. The space complexity is O(1) (excluding the input storage).

If the problem asks for the maximum number of non-overlapping intervals, simply return count instead of len(intervals)-count.

Answer: The minimum number of intervals to remove is given by the function above. For the sample input, the result is 1. (Adjust based on the actual input in the image.)
If the image refers to LeetCode 435, the code is correct. The final answer would be the output of this function for the given intervals. For example, if the input is [[1,2],[2,3],[3,4],[1,3]], the answer is 1.

Assuming the image's problem is LeetCode 435, the answer is the code provided, and the result for typical inputs is as explained.

Final Answer: \boxed{1} (example value; adjust based on actual input). For the standard test case, the answer is 1. So \boxed{1} is a common result.

\boxed{1}

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