Introduction

In this blog post, we will explore how to solve the LeetCode problem "2406" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 <= 106

Explanation

Here's a breakdown of the solution, followed by the code:

• Sort by Start Time: The most crucial step is to sort the intervals based on their start times. This allows us to process them in the order they appear on the number line.
• Priority Queue (Min-Heap): We use a min-heap (priority queue) to track the end times of the intervals currently assigned to groups. When considering a new interval, if its start time is greater than or equal to the earliest ending time in the heap, it means we can reuse that group. Otherwise, we need to create a new group.

• Counting Groups: The size of the heap at any point represents the number of active groups needed to accommodate the intervals processed so far. The maximum size the heap reaches is the minimum number of groups required.

• Time Complexity: O(N log N), where N is the number of intervals (due to sorting and heap operations). Space Complexity: O(N) in the worst case, to store the heap.

Code

    import heapq

def minGroups(intervals):
    """
    Finds the minimum number of groups needed to divide the intervals such that no two intervals in the same group intersect.

    Args:
        intervals: A list of lists, where each inner list represents an interval [start, end].

    Returns:
        The minimum number of groups needed.
    """

    # Sort the intervals based on their start times
    intervals.sort(key=lambda x: x[0])

    # Use a min-heap to track the end times of intervals in groups
    heap = []
    num_groups = 0

    for interval in intervals:
        start, end = interval

        # If the current interval's start time is greater than or equal to the earliest ending time,
        # we can reuse an existing group.
        if heap and start > heap[0]:
            heapq.heappop(heap)

        # Otherwise, we need to create a new group.
        heapq.heappush(heap, end)
        num_groups = len(heap)

    return num_groups