README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2400-2499/2406.Divide%20Intervals%20Into%20Minimum%20Number%20of%20Groups/README_EN.md	1713	Weekly Contest 310 Q3	Greedy Array Two Pointers Prefix Sum Sorting Heap (Priority Queue)

2406. Divide Intervals Into Minimum Number of Groups

中文文档

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 <= 105
• intervals[i].length == 2
• 1 <= lefti <= righti <= 106

Solutions

Solution 1: Greedy + Priority Queue (Min Heap)

First, we sort the intervals by their left endpoints. We use a min heap to maintain the rightmost endpoint of each group (the top of the heap is the minimum of the rightmost endpoints of all groups).

Next, we traverse each interval:

• If the left endpoint of the current interval is greater than the top element of the heap, it means the current interval can be added to the group where the top element of the heap is located. We directly pop the top element of the heap, and then put the right endpoint of the current interval into the heap.
• Otherwise, it means there is currently no group that can accommodate the current interval, so we create a new group and put the right endpoint of the current interval into the heap.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array intervals.

Python3

class Solution:
    def minGroups(self, intervals: List[List[int]]) -> int:
        q = []
        for left, right in sorted(intervals):
            if q and q[0] < left:
                heappop(q)
            heappush(q, right)
        return len(q)

Java

class Solution {
    public int minGroups(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        PriorityQueue<Integer> q = new PriorityQueue<>();
        for (var e : intervals) {
            if (!q.isEmpty() && q.peek() < e[0]) {
                q.poll();
            }
            q.offer(e[1]);
        }
        return q.size();
    }
}

C++

class Solution {
public:
    int minGroups(vector<vector<int>>& intervals) {
        sort(intervals.begin(), intervals.end());
        priority_queue<int, vector<int>, greater<int>> q;
        for (auto& e : intervals) {
            if (q.size() && q.top() < e[0]) {
                q.pop();
            }
            q.push(e[1]);
        }
        return q.size();
    }
};

Go

func minGroups(intervals [][]int) int {
	sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] < intervals[j][0] })
	q := hp{}
	for _, e := range intervals {
		if q.Len() > 0 && q.IntSlice[0] < e[0] {
			heap.Pop(&q)
		}
		heap.Push(&q, e[1])
	}
	return q.Len()
}

type hp struct{ sort.IntSlice }

func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}

TypeScript

function minGroups(intervals: number[][]): number {
    intervals.sort((a, b) => a[0] - b[0]);
    const q = new PriorityQueue({ compare: (a, b) => a - b });
    for (const [left, right] of intervals) {
        if (!q.isEmpty() && q.front() < left) {
            q.dequeue();
        }
        q.enqueue(right);
    }
    return q.size();
}