Python three step process:  sort, segment and count
Intuition\n Describe your first thoughts on how to solve this problem. \nThe "such that"s of the problem description -- overlapping ranges go into the same grou
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe "such that"s of the problem description -- overlapping ranges go into the same group and no range is in more than one group -- suggested to me that we think of another construct that I called "segments", where a segment is a maximal collection of ranges that overlap each other somehow.  With this, we can reduce the problem to splitting "segments" rather than ranges, since that will satisfy the "such that"s.  We can think about the solution as having three parts:   First, sort the ranges by their "low" value.  Second, partition the sorted ranges into segments.  Third, count how many ways we can group the segments.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nAfter sorting, we run through the sorted ranges, keeping track of how many ranges belong to the "current segment".  This can be done handily by max-ing with the "high" value of each range we process to maintain the highest value in the current segment.  As we go through the sorted ranges, we ask whether each overlaps with the current segment "high value".  If so, we increment the range count of the current segment and possibly update the segment high value.  If not, i.e., the next range doesn\'t overlap with the current segment, we note that we\'ve finished the current segment and record its size in a list.  In fact, the size doesn\'t matter, but it\'s nice to have a list of *something* associated with each segment.  The punch line is:  we end up with a *number of segments*.\n\nThe third step is to count how many ways we could split the segments into two groups, which we can translate into answering the question "which group should this go into" for each segment.  It\'s a short hop to observing that the number of ways of answering the question for all the segments is 2 ^ n_segments, where n_segments is the length of our list from the second step.  Now, if "n_segments" is small, say, if it\'s 3, then calculating 2 ^ n_segments is easy -- ```2 ** n_segments```.   BUT (there\'s always a BUT) if none of the ranges overlap, then n_segments can be equal to the number of ranges, which can be 10 ** 5.  2 ** ( 10 ** 5 ) is a big number (!), which is why the question asks us to report the result modulo 1000000007, a convenient prime number that fits neatly into a 32 bit signed integer.\n\nThe way around this is a little hard to explain, but here\'s the general idea:  Think of n_segments as a binary number, where each bit represents multiplying (or not multiplying if the bit is 0) the result by a power of two.  The trick is:  as we\'re doing the multiplication of our result-so-far by a power of two, we do it modulo 1000000007, and when we incrementally update the power of two by squaring it, we also take the result modulo 1000000007.  This means that we never need to deal with a number bigger than 1000000007 ** 2, which conveniently fits into a 64 bit signed ```long```.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nNotwithstanding all the verbiage above, the steps of segmenting the ranges and then evaluating the exponential expression take $$O(n)$$ and $$O(log n)$$ respectively.  However, we also include the up-front sort, which is $$O(n log n)$$ and that ends up defining the overall time complexity.\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe only non-scalar data we use (besides the input) is the list of segment sizes, which can grow as $$O(n)$$.  However, as I mentioned above, we can replace this with a scalar count of the number of segments, so (if I weren\'t too lazy to change the code) we could reduce the space complexity to $$O(1)$$.\n# Code\n```\nimport math\nclass Solution:\n    def countWays(self, ranges: list[list[int]]) -> int:\n        # Each "range" of integers is defined by two numbers [low, high] that\n        # are inclusive.\n\n        # First, sort "ranges" by their "low" values:\n        ranges.sort()    # Thanks, Python!\n\n        # Next, go through the sorted "ranges"" to partition them into segments\n        # consisting of overlapping subsets of "ranges" and simply count how\n        # many ranges are in each segment:\n        segments = []\n        # Keep track of largest "high" value in current segment, so it\'s easy to\n        # tell whether each range we encounter belongs in the current segment or\n        # is the start of a new segment\n        segment_high = -math.inf\n        segment_count = 0\n        for low, high in ranges:\n            if segment_high < low:   # No overlap, so save current segment and start new one\n                if segment_count > 0:\n                    segments.append( segment_count )\n                segment_high = high\n                segment_count = 1\n            else:   # Overlap, so range belongs to current segment:\n                segment_high = max( segment_high, high )\n                segment_count = 1\n        if segment_count > 0:\n            segments.append( segment_count )\n        # OK, so now we have a list of segment_counts.  Now we have to split them.  We\'ve\n        # already taken care of both of the "such that"s by segmentizing.  The splitting\n        # consists of assigning segments into one of two groups.  For this, we merely\n        # have to associate with each segment a label 0 or 1 depending on which group\n        # it\'s assigned.  So, we have 2 ^ n_segments ways to split.   Since we could\n        # have up to "n_ranges" segments, which can range up to 10^5, we have to be\n        # careful about evaluating the exponential expression.  This uses the approach\n        # that we take "n_segments" and break it down into terms 2^k * bit_k_of_n_ranges\n        # and incrementally evaluate 2^k:\n        MODULUS = 1000000007\n        power_of_two = 2\n        count = 1\n        n_segments = len( segments )\n        while n_segments:\n            bit = 1 & n_segments\n            if bit:\n                count = ( count * power_of_two ) % MODULUS\n            power_of_two = ( power_of_two * power_of_two ) % MODULUS\n            n_segments >>= 1\n        return count\n\n```
['Python3']