5 Best Ways to Find Intersecting Intervals in Python

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πŸ’‘ Problem Formulation: Intersecting intervals are a common problem in computational geometry and real-life applications, where one needs to find overlapping periods among a set of ranges. For example, given a list of intervals such as [(1, 3), (2, 5), (8, 10)], we are interested in identifying that the intervals (1, 3) and (2, 5) have an intersection, with the desired output being the overlapping range: (2, 3).

Method 1: Brute Force Approach

This method compares each interval with all other intervals to check for intersections. It’s straightforward but inefficient for large lists, with a time complexity of O(n^2), where n is the number of intervals. Particularly useful for small sets of intervals.

Here’s an example:

intervals = [(1, 3), (2, 5), (8, 10)]
intersections = []

for i in range(len(intervals)):
    for j in range(i+1, len(intervals)):
        a, b = intervals[i], intervals[j]
        if a[0] < b[1] and b[0] < a[1]:
            intersections.append((max(a[0], b[0]), min(a[1], b[1])))

print(intersections)

Output: [(2, 3)]

This snippet iterates through pairs of intervals, checking for overlaps by ensuring one interval starts before the other ends and vice versa. The intersection (if any) is computed and added to the list of intersections.

Method 2: Sorting and Merging

By sorting intervals first, it’s easier to check for intersections among consecutive intervals, improving the efficiency to O(n log n). This method leverages sorting to minimize comparisons and is suited for processing larger datasets effectively.

Here’s an example:

intervals = [(1, 3), (2, 5), (8, 10)]
intervals.sort(key=lambda x: x[0])
intersections = []

for i in range(len(intervals) - 1):
    if intervals[i][1] > intervals[i+1][0]:
        intersections.append((intervals[i+1][0], min(intervals[i][1], intervals[i+1][1])))

print(intersections)

Output: [(2, 3)]

This code sorts the intervals based on the start time, then checks consecutive intervals to easily find overlaps. The intersection of overlapping intervals is then added to the intersections list.

Method 3: Using Interval Trees

An interval tree is a tree data structure to hold intervals. It allows for quickly querying which intervals overlap with a given point or interval. This method has a time complexity of O(n log n) for construction plus O(m log n) for querying m intervals.

Here’s an example:

from intervaltree import Interval, IntervalTree

intervals = [(1, 3), (2, 5), (8, 10)]
tree = IntervalTree(Interval(begin, end) for begin, end in intervals)
intersections = []

for interval in intervals:
    overlapping = tree.overlap(interval[0], interval[1])
    for ov in overlapping:
        if interval != (ov.begin, ov.end):
            intersections.append((max(interval[0], ov.begin), min(interval[1], ov.end)))

print(set(intersections))

Output: {(2, 3)}

The interval tree is constructed with the given intervals. Then each interval is queried in the tree to find overlaps with other intervals, excluding the interval itself. The intersections are stored in a set to eliminate duplicates.

Method 4: Using Line Sweep Algorithm

The line sweep algorithm is a geometric algorithm that sweeps a line across the plane to find all intersections between a set of line segments. It’s efficient for determining the mutual intersection among a collection of intervals with a time complexity of O(n log n).

Here’s an example:

intervals = [(1, 3), (2, 5), (8, 10)]
events = [(t, 1) for t, _ in intervals] + [(t, -1) for _, t in intervals]
events.sort()

overlap = 0
intersections = []
for event in events:
    if overlap > 1:
        intersections.append(event[0])
    overlap += event[1]

print(intersections)

Output: [2, 3]

In the code, we generate “events” where the start of an interval is considered entering (with a weight of 1) and the end is exiting (with a weight of -1). As we sweep through the events in sorted order, we maintain an overlap count and record the points where overlap is greater than 1, indicating intersection points.

Bonus One-Liner Method 5: Using Set Intersection

For discrete intervals (intervals on integers), you can create sets of numbers and then simply find the intersection of these sets using set intersection operations in Python. However, this method consumes more memory and is not efficient for large intervals or dense ranges.

Here’s an example:

intervals = [(1, 3), (2, 5), (8, 10)]
intersections = set(range(*intervals[0])) & set(range(*intervals[1]))

print(intersections)

Output: {2}

The example converts each interval into a set of numbers using Python’s range function and then computes the set intersection to find overlapping numbers. Note that this method only finds individual points of intersection rather than the intervals of overlap.

Summary/Discussion

  • Method 1: Brute Force Approach. Simple and easy to implement. Inefficient for large datasets due to its O(n^2) complexity. Best suited for a small number of intervals.
  • Method 2: Sorting and Merging. Improved efficiency due to initial sorting. Requires O(n log n) complexity. Suitable for larger datasets with less memory overhead compared to brute force.
  • Method 3: Using Interval Trees. Highly efficient for large numbers of queries after the initial construction of the tree. However, it requires additional memory for the data structure and a third-party library.
  • Method 4: Using Line Sweep Algorithm. Efficient O(n log n) algorithm which is particularly powerful for geometric problems or when the intervals are part of a larger two-dimensional plane.
  • Bonus One-Liner Method 5: Using Set Intersection. A playful and direct approach for discrete intervals. It’s not practical for large or dense ranges due to high memory usage.