leetcode - 2406. Divide Intervals Into Minimum Number of Groups

Description

You are given a 2D integer array intervals where intervals[i] = [lefti, righti] represents the inclusive interval [lefti, righti].

You have to divide the intervals into one or more groups such that each interval is in exactly one group, and no two intervals that are in the same group intersect each other.

Return the minimum number of groups you need to make.

Two intervals intersect if there is at least one common number between them. For example, the intervals [1, 5] and [5, 8] intersect.

Example 1:

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.

Example 2:

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.

Constraints:

1 <= intervals.length <= 10^5
intervals[i].length == 2
1 <= lefti <= righti <= 10^6

Solution

Greedy, sort by start, then for each one, check if it could fit into existing groups, if so, then fit, else create a new group for it.

Because we only need to check the end of existing intervals for fitting, so we use a min heap. If the new interval could fit in the earliest ending group, then fit (pop from heap and push new number). If not, then create a new group (push new number into heap)

Time complexity: $o(\log n)$
Space complexity: $o(n)$

Code

class Solution:
    def minGroups(self, intervals: List[List[int]]) -> int:
        intervals.sort(key=lambda x:x[0])
        heap = []
        heapq.heappush(heap, intervals[0][1])
        for i in range(1, len(intervals)):
            if heap[0] < intervals[i][0]:
               heapq.heappop(heap)
               heapq.heappush(heap, intervals[i][1])
            else:
                heapq.heappush(heap, intervals[i][1])
        return len(heap)