2580. Count Ways to Group Overlapping Ranges - Detailed Explanation

Problem Statement

You are given a 2D integer array ranges where each element is a pair [l, r] representing an interval on the number line. Two ranges are said to be overlapping if they share at least one common point. In other words, range [l₁, r₁] overlaps with [l₂, r₂] if and only if
l₂ ≤ r₁ and l₁ ≤ r₂.

You want to partition the given ranges into one or more groups subject to the following rule:

• If two ranges overlap, they must be in the same group.
• For ranges that do not overlap, you have the freedom to either put them into the same group or separate them into different groups.

Your task is to count the total number of ways to form such groups. Since the answer can be large, return it modulo 10⁹ + 7.

Example 1

• Input: ranges = [[1, 3], [5, 7], [8, 10]], where k is not needed explicitly
• Explanation:
  • Ranges [1, 3] and [5, 7] do not overlap (since 3 < 5). Similarly, [5, 7] and [8, 10] do not overlap (7 < 8).
  • There are two “gaps” between consecutive ranges. For each gap, you have a binary choice:
    • Either keep the ranges in separate groups,
    • Or merge them with the previous group.
  • Thus, the number of ways = 2² = 4.
• Output: 4

Example 2

• Input: ranges = [[1, 5], [2, 6], [8, 10]]
• Explanation:
  • Ranges [1, 5] and [2, 6] overlap (they share common points), so they must be in the same group.
  • The range [8, 10] does not overlap with the merged group ([1, 5] ∪ [2, 6]).
  • There is one gap (between the merged group and [8, 10]).
  • Number of ways = 2¹ = 2.
• Output: 2

Example 3

• Input: ranges = [[3, 4], [5, 6], [7, 8]]
• Explanation:
  • None of the ranges overlap, so between each adjacent pair there is a gap.
  • With 3 ranges, there are 2 gaps.
  • Number of ways = 2² = 4.
• Output: 4

Hints

1. Overlapping Implies Forced Grouping:
   If two ranges overlap, they must be grouped together. This forces merging of overlapping intervals into one component.

2. Non-Overlapping Provides Flexibility:
   When two consecutive ranges (after sorting) do not overlap, you can decide either to merge them into the same group or leave them as separate groups. Each such gap gives you a binary decision.

3. Sorting is the Key:
   Sorting the ranges by their starting point allows you to “sweep” the number line from left to right and detect gaps (non-overlapping adjacencies) easily.

Approach 1: Gap Counting After Sorting

Idea

1. Sort the Ranges:
   First, sort the given ranges by their starting point. This makes it easier to compare consecutive ranges and check for overlaps.

2. Merge Overlapping Intervals:
   As you iterate over the sorted list, maintain a variable (say current_end) that keeps track of the farthest right endpoint of the current merged group of overlapping intervals.

3. Count Gaps:

   • For each new range, if its start is greater than the current_end, it does not overlap with the previous group.
   • Every time this occurs, you have a “gap” where you have a binary choice: either join this range with the previous group (i.e., “merge” despite the gap) or let it form a new group.
   • Count the total number of such gaps.

4. Calculate the Answer:
   If there are g gaps, the total number of valid groupings is:
   Answer = 2^(g) mod (10⁹ + 7)

Visual Walkthrough

Consider ranges = [[1, 3], [5, 7], [8, 10]]:

1. Sort the Ranges:
   The list is already sorted.

2. Initialize:

   • Set current_end = 3 (from the first range).
   • Set gaps = 0.

3. Iterate:

   • For the second range [5, 7]:
     • Check if 5 > current_end (3). It is, so there is a gap. Increase gaps to 1.
     • Update current_end to 7.
   • For the third range [8, 10]:
     • Check if 8 > current_end (7). It is, so increase gaps to 2.
     • Update current_end to 10.

4. Calculate Ways:

   • There are 2 gaps, so answer = 2² = 4.

Code Solutions

Python Code