Published: Sep 21, 2022

## Problem Description

You are given a 2D integer array

`intervals`

where`intervals[i] = [left[i], right[i]]`

represents the inclusive interval`[left[i], right[i]]`

.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.Constraints:

`1 <= intervals.length <= 10**5`

`intervals[i].length == 2`

`1 <= left[i] <= right[i] <= 10**6`

https://leetcode.com/problems/divide-intervals-into-minimum-number-of-groups/

## Examples

```
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.
```

## How to Solve

Interval problems have two well-used approaches. One starts from sorting the given array and uses heap for comparison. This type of approach is common to merging intervals problems. Another counts plus/minus for start/end. This approach is often useful to solve meeting room usage problems.

The solution here took the second, plus/minus, approach. To divide into groups, we should find the maximum duplicates. That will be the same as the minimum number of groups. To save memory space, room usage counts are saved in a Hash Table except Java and JavaScript solution. Java’s sorted map, TreeMap, is not so fast compared to C++’s map. In JavaScript, sorting a map is not so fast. For those reasons, Java and JavaScript use int array of enough size. For each interval, count up start time and count down end + 1 time. Then, create a prefix sum ordered by the time. The maximum value in the prefix sum is the answer.

For a reference, Python solution has the first approach, sorting and heap.

## Solution

```
#include <map>
#include <vector>
using namespace std;
class DivideIntervalsIntoMinimumNumberOfGroups {
public:
int minGroups(vector<vector<int>>& intervals) {
map<int, int> rooms;
for (int i = 0; i < intervals.size(); ++i) {
rooms[intervals[i][0]]++;
rooms[intervals[i][1] + 1]--;
}
int result = 0, cur = 0;
for (auto &[key, value] : rooms) {
cur += value;
result = max(result, cur);
}
return result;
}
};
```

```
import java.util.*;
class DivideIntervalsIntoMinimumNumberOfGroups {
public int minGroups(int[][] intervals) {
int n = 1000002;
int[] rooms = new int[n];
for (int i = 0; i < intervals.length; ++i) {
rooms[intervals[i][0]]++;
rooms[intervals[i][1] + 1]--;
}
int result = 0;
for (int i = 1; i < n; ++i) {
rooms[i] += rooms[i - 1];
result = Math.max(result, rooms[i]);
}
return result;
}
}
```

```
/**
* @param {number[][]} intervals
* @return {number}
*/
var minGroups = function(intervals) {
const n = 1000002;
let rooms = new Array(n).fill(0);
for (let [s, e] of intervals) {
rooms[s]++;
rooms[e + 1]--;
}
let result = 0;
for (let i = 1; i < n; i++) {
rooms[i] += rooms[i - 1];
result = Math.max(result, rooms[i]);
}
return result;
};
```

```
class DivideIntervalsIntoMinimumNumberOfGroups:
def minGroups(self, intervals: List[List[int]]) -> int:
rooms = collections.defaultdict(int)
for s, e in intervals:
rooms[s] += 1
rooms[e + 1] -= 1
result, cur = 0, 0
for t in sorted(rooms):
cur += rooms[t]
result = max(result, cur)
return result
def minGroupsByHeap(self, intervals: List[List[int]]) -> int:
intervals.sort()
q = []
for s, e in intervals:
if q and q[0] < s:
heapq.heappop(q)
heapq.heappush(q, e)
return len(q)
```

```
# @param {Integer[][]} intervals
# @return {Integer}
def min_groups(intervals)
rooms = {}
intervals.each do |s, e|
rooms[s] = (rooms[s] || 0) + 1
rooms[e + 1] = (rooms[e + 1] || 0) - 1
end
result, cur = 0, 0
rooms.sort.each do |_, value|
cur += value
result = [result, cur].max
end
result
end
```

## Complexities

- Time:
`O(n + m)`

– n: number of intervals, m: number of unique start/end times - Space:
`O(m)`