Leetcode 2406: Divide Intervals Into Minimum Number of Groups
You are given a 2D array of intervals, where each interval is represented as a pair of integers [left, right], denoting an inclusive range. Your task is to divide these intervals into the minimum number of groups such that no two intervals within the same group overlap. Two intervals overlap if there is at least one common number between them.
Problem
Approach
Steps
Complexity
Input: The input consists of a 2D array `intervals` where each element `intervals[i]` is an array of two integers, left and right, representing the start and end of an interval.
Example: intervals = [[1, 5], [6, 10], [3, 7], [8, 12], [15, 18]]
Constraints:
• 1 <= intervals.length <= 10^5
• Each interval is represented by two integers where 1 <= lefti <= righti <= 10^6
Output: Return the minimum number of groups required to partition the intervals such that no two intervals within the same group overlap.
Example: Output: 3
Constraints:
Goal: The goal is to minimize the number of groups while ensuring that no two intervals within the same group overlap.
Steps:
• 1. Sort the intervals by their start times.
• 2. Use a priority queue (min-heap) to track the end times of intervals in each group.
• 3. For each interval, check if it can be added to an existing group (i.e., its start time is greater than the smallest end time in the heap).
• 4. If it can, update the group's end time; otherwise, start a new group.
• 5. Keep track of the maximum number of groups at any time.
Goal: The solution should be able to handle the maximum constraints efficiently.
Steps:
• The time complexity of the solution should be O(n log n), where n is the number of intervals.
Assumptions:
• Intervals are inclusive of both left and right endpoints.
• The input list will not contain empty intervals.
• Input: intervals = [[5, 10], [6, 8], [1, 5], [2, 3], [1, 10]]
• 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]. Thus, 3 groups are needed.
• Input: intervals = [[1, 3], [5, 6], [8, 10], [11, 13]]
• Explanation: Since none of the intervals overlap, all intervals can be placed in one group.
• Input: intervals = [[1, 4], [3, 5], [5, 6], [7, 8]]
• Explanation: The optimal partition is: Group 1: [1, 4], [5, 6]; Group 2: [3, 5], [7, 8]. We need 2 groups.
Approach: We will use a greedy approach, sorting the intervals and using a priority queue (min-heap) to track the current end times of intervals in each group. By efficiently managing the end times, we can minimize the number of groups.
Observations:
• Sorting the intervals will allow us to process them in order of their start times.
• Using a priority queue will allow us to efficiently manage and track the end times of intervals in each group.
• A greedy strategy works well here, as we only need to track the end times and adjust the groups accordingly.
Steps:
• 1. Sort the intervals based on their start time.
• 2. Initialize a priority queue and start processing the intervals.
• 3. For each interval, check if it can be added to the existing groups by comparing it with the smallest end time in the queue.
• 4. If the interval overlaps with an existing group, update the group's end time. Otherwise, create a new group by adding the interval to the priority queue.
• 5. Track the maximum number of groups required.
Empty Inputs:
• Empty input is not possible as per the problem constraints, but handling zero intervals would result in zero groups.
Large Inputs:
• For large inputs with up to 10^5 intervals, the solution should perform efficiently in O(n log n) time due to sorting and priority queue operations.
Special Values:
• Intervals with the same start and end time (e.g., [1, 1], [1, 1]) should be handled correctly as separate intervals.
Constraints:
• The solution must efficiently handle intervals where the length of the list is large (up to 10^5).
int minGroups(vector<vector<int>>& list) {
sort(list.begin(), list.end());
int res = 1, cnt = 1;
priority_queue<int, vector<int>, greater<int>> pq;
int n = list.size();
pq.push(list[0][1]);
if(n == 1) return 1;
for(int i = 1; i < n; i++) {
int next = list[i][0];
while(!pq.empty() && next > pq.top()) {
pq.pop();
cnt--;
}
pq.push(list[i][1]);
cnt++;
res = max(res, cnt);
}
return res;
}
1 : Function Definition
int minGroups(vector<vector<int>>& list) {
Define the function `minGroups` which takes a vector of vectors representing meeting time intervals and returns the minimum number of meeting groups.
2 : Sorting
sort(list.begin(), list.end());
Sort the meeting intervals based on their start times to facilitate easier management of overlapping intervals.
3 : Variable Declaration
int res = 1, cnt = 1;
Initialize the result variable `res` to 1 and the counter `cnt` to 1, representing the number of ongoing meetings.
4 : Priority Queue
priority_queue<int, vector<int>, greater<int>> pq;
Define a priority queue `pq` to store the end times of the ongoing meetings in ascending order.
5 : Size Calculation
int n = list.size();
Get the size of the list of meeting intervals, `n`, to determine the number of intervals.
6 : First Interval Push
pq.push(list[0][1]);
Push the end time of the first meeting interval into the priority queue.
7 : Early Return
if(n == 1) return 1;
If there is only one interval, return 1 as only one group is needed.
8 : Main Loop
for(int i = 1; i < n; i++) {
Start a loop to iterate over the remaining intervals starting from index 1.
9 : Next Start Time
int next = list[i][0];
Get the start time of the next meeting interval.
10 : While Loop
while(!pq.empty() && next > pq.top()) {
While the priority queue is not empty and the next start time is greater than the earliest ending meeting time, pop meetings from the queue.
11 : Priority Queue Pop
pq.pop();
Pop the earliest ending meeting time from the priority queue.
12 : Decrease Counter
cnt--;
Decrement the ongoing meetings count since one meeting has ended.
13 : Push End Time
pq.push(list[i][1]);
Push the end time of the current meeting interval into the priority queue.
14 : Increase Counter
cnt++;
Increment the ongoing meetings count as a new meeting has started.
15 : Update Maximum
res = max(res, cnt);
Update the result to keep track of the maximum number of overlapping meetings.
16 : Return Result
return res;
Return the result, which is the maximum number of overlapping meetings.
Best Case: O(n log n), where n is the number of intervals, due to sorting.
Average Case: O(n log n), as sorting dominates the time complexity.
Worst Case: O(n log n), as the sorting step is the most time-consuming operation.
Description: The solution efficiently handles large inputs due to the O(n log n) time complexity.
Best Case: O(n), as we always need space for storing the intervals and their end times.
Worst Case: O(n), where n is the number of intervals, as we store the intervals and the priority queue.
Description: The space complexity is linear due to the storage of intervals and the priority queue used to manage the end times.
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