How do you solve Divide Intervals Into Minimum Number of Groups on LeetCode?

Divide Intervals Into Minimum Number of Groups (LeetCode #2406) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: The problem is about managing overlapping intervals efficiently.

What is the time complexity of Divide Intervals Into Minimum Number of Groups?

The optimal solution for Divide Intervals Into Minimum Number of Groups runs in O(n log n) time and uses O(n) space. The sorting step takes O(n log n), and managing the heap takes O(n) in total, leading to an overall complexity of O(n log n).

What is the brute force solution for Divide Intervals Into Minimum Number of Groups?

The brute force approach for Divide Intervals Into Minimum Number of Groups is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible grouping of intervals to see if they intersect. This is simple but inefficient, as it requires checking all combinations. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Divide Intervals Into Minimum Number of Groups?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as completely overlapping intervals or completely disjoint intervals. Practice explaining your thought process while coding; it helps in interviews.

What are common mistakes when solving Divide Intervals Into Minimum Number of Groups?

Students often forget to sort the intervals before processing. Not using a priority queue effectively can lead to incorrect group counts.

#2406

Divide Intervals Into Minimum Number of Groups

Medium

ArrayTwo PointersGreedySortingHeap (Priority Queue)Prefix SumGreedySortingHeap (Priority Queue)

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal approach uses sorting and a priority queue to efficiently manage overlapping intervals. By tracking the end times of intervals, we can determine the maximum number of overlapping intervals at any point.

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Algorithm

3 steps

1Step 1: Sort the intervals based on their start times.
2Step 2: Use a priority queue (min-heap) to track the end times of the current groups.
3Step 3: For each interval, check if it can fit into an existing group (i.e., if its start time is greater than the smallest end time in the heap). If it can, replace that end time; otherwise, add a new group.

solution.py11 lines

1# Full working Python code
2import heapq
3
4def minGroups(intervals):
5    intervals.sort(key=lambda x: x[0])
6    min_heap = []
7    for interval in intervals:
8        if min_heap and min_heap[0] < interval[0]:
9            heapq.heappop(min_heap)
10        heapq.heappush(min_heap, interval[1])
11    return len(min_heap)

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Complexity note: The sorting step takes O(n log n), and managing the heap takes O(n) in total, leading to an overall complexity of O(n log n).

1The problem is about managing overlapping intervals efficiently.
2Using a priority queue helps in tracking the end times of intervals, allowing for quick decisions on grouping.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.