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

Code

解法同：My Calendar III，只需要反過來想，有多少重疊，就代表要分成多少個 group。

Time Complexity: $O(n\log n)$, Space Complexity: $O(n)$

class Solution {
public:
    int minGroups(vector<vector<int>>& intervals) {
        map<int, int> mp;
        for(auto interval: intervals) {
            mp[interval[0]]++;
            mp[interval[1] + 1]--;
        }
 
        int minGroup = 1, curGroup = 0;
        for(auto it: mp) {
            curGroup += it.second;
            minGroup = max(minGroup, curGroup);
        }
 
        return minGroup;
    }
};

Source

• Divide Intervals Into Minimum Number of Groups - LeetCode