Introduction

In this blog post, we will explore how to solve the LeetCode problem "2580" 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 ranges where ranges[i] = [starti, endi] denotes that all integers between starti and endi (both inclusive) are contained in the ith range. You are to split ranges into two (possibly empty) groups such that: Each range belongs to exactly one group. Any two overlapping ranges must belong to the same group. Two ranges are said to be overlapping if there exists at least one integer that is present in both ranges. For example, [1, 3] and [2, 5] are overlapping because 2 and 3 occur in both ranges. Return the total number of ways to split ranges into two groups. Since the answer may be very large, return it modulo 109 + 7. Example 1: Input: ranges = [[6,10],[5,15]] Output: 2 Explanation: The two ranges are overlapping, so they must be in the same group. Thus, there are two possible ways: - Put both the ranges together in group 1. - Put both the ranges together in group 2. Example 2: Input: ranges = [[1,3],[10,20],[2,5],[4,8]] Output: 4 Explanation: Ranges [1,3], and [2,5] are overlapping. So, they must be in the same group. Again, ranges [2,5] and [4,8] are also overlapping. So, they must also be in the same group. Thus, there are four possible ways to group them: - All the ranges in group 1. - All the ranges in group 2. - Ranges [1,3], [2,5], and [4,8] in group 1 and [10,20] in group 2. - Ranges [1,3], [2,5], and [4,8] in group 2 and [10,20] in group 1. Constraints: 1 <= ranges.length <= 105 ranges[i].length == 2 0 <= starti <= endi <= 109

Explanation

Here's the breakdown of the solution:

• Merge Overlapping Ranges: The core idea is to first merge all overlapping ranges into a set of non-overlapping ranges. This simplifies the problem significantly.
• Count Independent Components: After merging, the non-overlapping ranges represent independent components. Each component can be placed in either group 1 or group 2.

• Calculate Power of 2: The number of ways to split the ranges is 2^{number of independent components}, since each independent component has two choices.

• Time Complexity: O(N log N), due to sorting.

• Space Complexity: O(N) in the worst case, where N is the number of ranges, for storing the merged ranges.

Code

    def count_ways_to_split(ranges):
    ranges.sort()
    merged_ranges = []
    for start, end in ranges:
        if not merged_ranges or start > merged_ranges[-1][1]:
            merged_ranges.append([start, end])
        else:
            merged_ranges[-1][1] = max(merged_ranges[-1][1], end)

    num_components = len(merged_ranges)

    MOD = 10**9 + 7
    result = pow(2, num_components, MOD)
    return result