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Count Ways to Group Overlapping Ranges

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IBM
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Topics:
ArraysGraphsGreedy Algorithms

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

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What are the constraints on the size of the `ranges` array, and the values of `starti` and `endi`?
  2. Can the ranges be empty (starti > endi)? What should I return in that case?
  3. If ranges are adjacent but not overlapping (e.g., [1, 2], [2, 3]), can they be in the same group?
  4. Can the ranges be unsorted in each inner list? Should I sort them?
  5. If there are no possible groups (i.e. every range overlaps with every other range), what should be returned?

Brute Force Solution

Approach

The brute force method for grouping overlapping ranges involves trying every possible grouping. We consider each range individually, deciding whether to group it with a prior group or start a new one. This exhaustive exploration guarantees we find all valid groupings, even if it's incredibly inefficient.

Here's how the algorithm would work step-by-step:

  1. Take the first range and consider it as its own group.
  2. Move to the second range, and try adding it to the first range's group, checking if they overlap.
  3. If they don't overlap, consider the second range as a separate new group.
  4. Now, move to the third range. Try adding it to the first group. Then, try adding it to the second group (if it exists). If it doesn't overlap with either, create a new group with just the third range.
  5. Continue this process for every range. At each step, try adding the current range to every existing group, and if it doesn't fit in any, make a new group.
  6. Each time you finish going through all ranges, count that particular arrangement as one possible solution.
  7. Repeat all steps above, but change the initial groupings. For example, on the next try, put the first and second range together if they overlap, and keep going from there.
  8. Do this until you've exhaustively tried every possible combination of groupings.
  9. Finally, count all the valid grouping arrangements you found during this exhaustive process.

Code Implementation

def count_grouping_ways_brute_force(ranges):
    number_of_ranges = len(ranges)
    count = 0

    def overlap(range1, range2):
        return not (range1[1] < range2[0] or range2[1] < range1[0])

    def solve(index, current_groups):
        nonlocal count

        # Base case: all ranges have been processed
        if index == number_of_ranges:
            count += 1
            return

        # Try adding the current range to existing groups
        for group_index in range(len(current_groups)):
            can_add_to_group = True
            for existing_range in current_groups[group_index]:
                if not overlap(ranges[index], existing_range):
                    continue
                else:
                    break
            else:
                # Create a new copy of the groups and add the current range.
                new_groups = [group[:] for group in current_groups]
                new_groups[group_index].append(ranges[index])
                solve(index + 1, new_groups)

        # Always consider starting a new group
        # This ensures that every possibility is explored
        new_groups = [group[:] for group in current_groups]
        new_groups.append([ranges[index]])
        solve(index + 1, new_groups)

    # Start with an empty list of groups.
    solve(0, [])
    return count

Big(O) Analysis

Time Complexity
O(2^n)The provided brute force approach explores every possible grouping of the n ranges. For each range, we have two choices: either add it to an existing group or start a new group. Since we have n ranges, this leads to 2 * 2 * ... * 2 (n times) possibilities, which is 2^n. Therefore, the algorithm effectively tries all 2^n possible groupings. Thus, the time complexity grows exponentially with the number of ranges.
Space Complexity
O(2^N)The algorithm explores all possible groupings of N ranges. At each step, it makes a decision to either add a range to an existing group or start a new one, leading to a branching factor of up to N (the number of existing groups). This generates a decision tree where, in the worst-case, each range has to decide whether to belong to each of the existing groups or form a new group. This tree can have up to 2^N leaves, each representing a potential grouping configuration. Storing the current configuration (which groups each range belongs to) requires space proportional to N. The implicit recursion depth, in the worst case, is also N, so at any given time, the stack may hold N frames. The overall space is dictated by the need to maintain up to 2^N configurations as it explores the groupings.

Optimal Solution

Approach

The key idea is to realize overlapping ranges can be merged into single, larger ranges. We can efficiently count the ways to group these merged ranges by understanding that each group represents a 'yes/no' decision: either it's in a group, or it isn't.

Here's how the algorithm would work step-by-step:

  1. First, sort all the ranges based on their starting points. This makes it easy to process them in order.
  2. Go through the sorted ranges and merge any ranges that overlap. Overlapping ranges should be combined into single, bigger ranges. After this step, you have a collection of non-overlapping ranges.
  3. Count how many of these non-overlapping ranges you now have. This number represents the number of independent decisions you need to make about grouping.
  4. For each independent range, you can either put it in a group or not. This means there are two choices for each range.
  5. Since each range has two choices, and the choices are independent, the total number of ways to group the ranges is 2 raised to the power of the number of ranges.

Code Implementation

def count_grouping_ways(ranges):
    ranges.sort(key=lambda interval: interval[0])

    merged_ranges = []
    for interval in ranges:
        if not merged_ranges or interval[0] > merged_ranges[-1][1]:
            merged_ranges.append(interval)
        else:
            merged_ranges[-1] = [merged_ranges[-1][0], max(merged_ranges[-1][1], interval[1])]

    # The number of independent ranges determines the exponent for 2.
    number_of_independent_ranges = len(merged_ranges)

    #Calculate the total number of ways to group the ranges.
    total_ways = pow(2, number_of_independent_ranges)

    return total_ways

Big(O) Analysis

Time Complexity
O(n log n)The dominant time complexity comes from sorting the ranges initially, which takes O(n log n) time, where n is the number of ranges. The merging process involves iterating through the sorted ranges once, taking O(n) time. Counting the non-overlapping ranges also takes O(n) time. Therefore, the overall time complexity is dominated by the sorting step, resulting in O(n log n).
Space Complexity
O(N)The auxiliary space is dominated by the merged ranges. In the worst-case scenario, where no ranges overlap, we would need to store all N input ranges as non-overlapping ranges. Thus, the algorithm creates a new list of size N to store these merged ranges. Therefore the space complexity is O(N).

Edge Cases

CaseHow to Handle
ranges is null or emptyReturn 1, representing an empty grouping which is one possible way.
ranges contains only one rangeReturn 2, representing either including the range in a group or not including it (empty group).
All ranges are non-overlappingThe number of ways to group is 2 raised to the power of the number of ranges; calculate this using modular exponentiation.
All ranges are completely overlapping (e.g., [[0, 5], [1, 4], [2, 3]])The number of ways to group is 2 since all ranges must be in the same group or all not in any group.
ranges contains very large numbers, potentially causing integer overflowUse long data type where necessary to prevent overflow, especially during calculations such as modular exponentiation, and always take the modulo at each step.
ranges is extremely large (many ranges)Ensure the algorithm has a time complexity of O(n log n) or better by using efficient sorting and merging techniques.
ranges contains negative numbersThe algorithm should handle negative numbers correctly, especially when sorting and comparing ranges.
ranges contains overlapping ranges that are identical (e.g., [[1, 5], [1, 5]])Treat the duplicates as distinct ranges when calculating the number of possible groupings.