Divide Intervals Into Minimum Number of Groups

lcpy gen -n 2406   # by problem number
lcpy gen -s divide_intervals_into_minimum_number_of_groups   # by problem name

Input: intervals = [[5,10],[6,8],[1,5],[2,3],[1,10]]
Output: 3
Explanation: We can divide the intervals into the following groups:
- Group 1: [1, 5], [6, 8].
- Group 2: [2, 3], [5, 10].
- Group 3: [1, 10].
It can be proven that it is not possible to divide the intervals into fewer than 3 groups.

Input: intervals = [[1,3],[5,6],[8,10],[11,13]]
Output: 1
Explanation: None of the intervals overlap, so we can put all of them in one group.

class Solution:
    # Time: O(n log n)
    # Space: O(n)
    def min_groups(self, intervals: list[list[int]]) -> int:
        starts = sorted(left for left, _ in intervals)
        ends = sorted(r for _, r in intervals)
        groups = 0
        j = 0
        for start in starts:
            if start > ends[j]:
                j += 1
            else:
                groups += 1
        return groups