To remove a minimal number of intervals from the list such that the intervals that are left do not overlap, O(n*log n) algorithm exists:
def maximize_nonoverlapping_count(intervals):
# sort by the end-point
L = sorted(intervals, key=lambda (start, end): (end, (end - start)),
reverse=True) # O(n*logn)
iv = build_interval_tree(intervals) # O(n*log n)
result = []
while L: # until there are intervals left to consider
# pop the interval with the smallest end-point, keep it in the result
result.append(L.pop()) # O(1)
# remove intervals that overlap with the popped interval
overlapping_intervals = iv.pop(result[-1]) # O(log n + m)
remove(overlapping_intervals, from_=L)
return result
It should produce the following results:
f = maximize_nonoverlapping_count
assert f([[0, 133], [78, 100], [25, 30]]) == [[25, 30], [78, 100]]
assert f([[0,100],[9,10],[12,90]]) == [[9,10], [12, 90]]
assert f([[0, 100], [4, 20], [30, 35], [30, 78]]) == [[4, 20], [30, 35]]
assert f([[30, 70], [25, 40]]) == [[25, 40]]
It requires the data structure that can find in O(log n + m) time all intervals that overlap with the given interval e.g., IntervalTree. There are implementations that can be used from Python e.g., quicksect.py, see Fast interval intersection methodologies for the example usage.
Here's a quicksect-based O(n**2) implementation of the above algorithm:
from quicksect import IntervalNode
class Interval(object):
def __init__(self, start, end):
self.start = start
self.end = end
self.removed = False
def maximize_nonoverlapping_count(intervals):
intervals = [Interval(start, end) for start, end in intervals]
# sort by the end-point
intervals.sort(key=lambda x: (x.end, (x.end - x.start))) # O(n*log n)
tree = build_interval_tree(intervals) # O(n*log n)
result = []
for smallest in intervals: # O(n) (without the loop body)
# pop the interval with the smallest end-point, keep it in the result
if smallest.removed:
continue # skip removed nodes
smallest.removed = True
result.append([smallest.start, smallest.end]) # O(1)
# remove (mark) intervals that overlap with the popped interval
tree.intersect(smallest.start, smallest.end, # O(log n + m)
lambda x: setattr(x.other, 'removed', True))
return result
def build_interval_tree(intervals):
root = IntervalNode(intervals[0].start, intervals[0].end,
other=intervals[0])
return reduce(lambda tree, x: tree.insert(x.start, x.end, other=x),
intervals[1:], root)
Note: the time complexity in the worst case is O(n**2) for this implementation because the intervals are only marked as removed e.g., imagine such input intervals that len(result) == len(intervals) / 3 and there were len(intervals) / 2 intervals that span the whole range then tree.intersect() would be called n/3 times and each call would execute x.other.removed = True at least n/2 times i.e., n*n/6 operations in total:
n = 6
intervals = [[0, 100], [0, 100], [0, 100], [0, 10], [10, 20], [15, 40]])
result = [[0, 10], [10, 20]]
Here's a banyan-based O(n log n) implementation:
from banyan import SortedSet, OverlappingIntervalsUpdator # pip install banyan
def maximize_nonoverlapping_count(intervals):
# sort by the end-point O(n log n)
sorted_intervals = SortedSet(intervals,
key=lambda (start, end): (end, (end - start)))
# build "interval" tree O(n log n)
tree = SortedSet(intervals, updator=OverlappingIntervalsUpdator)
result = []
while sorted_intervals: # until there are intervals left to consider
# pop the interval with the smallest end-point, keep it in the result
result.append(sorted_intervals.pop()) # O(log n)
# remove intervals that overlap with the popped interval
overlapping_intervals = tree.overlap(result[-1]) # O(m log n)
tree -= overlapping_intervals # O(m log n)
sorted_intervals -= overlapping_intervals # O(m log n)
return result
Note: this implementation considers [0, 10] and [10, 20] intervals to be overlapping:
f = maximize_nonoverlapping_count
assert f([[0, 100], [0, 10], [11, 20], [15, 40]]) == [[0, 10] ,[11, 20]]
assert f([[0, 100], [0, 10], [10, 20], [15, 40]]) == [[0, 10] ,[15, 40]]
sorted_intervals and tree can be merged:
from banyan import SortedSet, OverlappingIntervalsUpdator # pip install banyan
def maximize_nonoverlapping_count(intervals):
# build "interval" tree sorted by the end-point O(n log n)
tree = SortedSet(intervals, key=lambda (start, end): (end, (end - start)),
updator=OverlappingIntervalsUpdator)
result = []
while tree: # until there are intervals left to consider
# pop the interval with the smallest end-point, keep it in the result
result.append(tree.pop()) # O(log n)
# remove intervals that overlap with the popped interval
overlapping_intervals = tree.overlap(result[-1]) # O(m log n)
tree -= overlapping_intervals # O(m log n)
return result
