2406 - Divide Intervals Into Minimum Number of Groups (Medium)

Problem Statement

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 <= 10^6

Approach 1: Line Sweep

Written by @wingkwong

class Solution {
public:
    // it's almost same as 253. Meeting Rooms II ...
    // the idea is to use line sweep to find out the number of overlapped intervals
    // if it is overlapped, that means you need to create a new group
    // e.g. if 2 intervals are overlapped, you need 2 groups ...
    // e.g. if 3 intervals are overlapped, you need 3 groups ...
    int minGroups(vector<vector<int>>& intervals) {
        int ans = 0, cnt = 0;
        // use map for internally sorting
        map<int, int> m;
        // standard line sweep
        // - increase the count of starting point by 1
        // - decrease the count of ending point by 1
        // - take prefix sum and do something
        for (auto& x: intervals) {
            // in - increase by 1
            m[x[0]]++;
            // out - decrease by 1
            m[x[1] + 1]--;
        }
        // so now what we have is
        // intervals   1  2  3  4  5  6  7  8  9  10
        // +           2  1  0  0  1  1  0  0  0  0 
        // -           0  0  1  0  1  0  0  1  0  2
        // m           2  1  -1 0  0  1  0  -1 0  -2
        for (auto& x: m) {
            // here we calculate the prefix sum
            cnt += x.second;
            // and record the maximum overlapping intervals
            ans = max(ans, cnt);
        }
        return ans;
    }
};