Introduction

In this blog post, we will explore how to solve the LeetCode problem "757" 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] = [starti, endi] represents all the integers from starti to endi inclusively. A containing set is an array nums where each interval from intervals has at least two integers in nums. For example, if intervals = [[1,3], [3,7], [8,9]], then [1,2,4,7,8,9] and [2,3,4,8,9] are containing sets. Return the minimum possible size of a containing set. Example 1: Input: intervals = [[1,3],[3,7],[8,9]] Output: 5 Explanation: let nums = [2, 3, 4, 8, 9]. It can be shown that there cannot be any containing array of size 4. Example 2: Input: intervals = [[1,3],[1,4],[2,5],[3,5]] Output: 3 Explanation: let nums = [2, 3, 4]. It can be shown that there cannot be any containing array of size 2. Example 3: Input: intervals = [[1,2],[2,3],[2,4],[4,5]] Output: 5 Explanation: let nums = [1, 2, 3, 4, 5]. It can be shown that there cannot be any containing array of size 4. Constraints: 1 <= intervals.length <= 3000 intervals[i].length == 2 0 <= starti < endi <= 108

Explanation

Here's the solution to the problem, with explanations:

• Sort Intervals: Sort the intervals based on their end points in ascending order. This greedy approach ensures that we prioritize intervals that end sooner, allowing us to maximize the overlap of the chosen numbers.

• Greedy Selection: Iterate through the sorted intervals. For each interval, check if it already contains at least two numbers in our 'containing set'. If not, greedily add the two largest possible numbers within that interval to the set. Choosing the largest possible numbers minimizes the chance of needing to add more numbers later for overlapping intervals.

• Runtime and Storage Complexity:

  • Runtime: O(N log N) due to sorting, where N is the number of intervals.
  • Storage: O(N) in the worst case for storing the 'containing set' of selected numbers.

Code

    def min_transfers(intervals):
    """
    Calculates the minimum possible size of a containing set.

    Args:
        intervals: A 2D integer array where intervals[i] = [starti, endi].

    Returns:
        The minimum possible size of a containing set.
    """

    intervals.sort(key=lambda x: x[1])  # Sort by end points
    selected = []  # The 'containing set' of numbers
    count = 0  # Count of numbers in the set

    for interval in intervals:
        start, end = interval
        needed = 2
        for num in selected:
            if start <= num <= end:
                needed -= 1

        if needed > 0:
            if needed == 1:
                val = end
                if val not in selected:
                    selected.append(val)
                    count+=1
            else:
                val1 = end - 1
                val2 = end
                if val1 not in selected:
                    selected.append(val1)
                    count+=1
                if val2 not in selected:
                    selected.append(val2)
                    count+=1


    return count