5 Best Ways to Find and Sort Overlapping Intervals in Python

💡 Problem Formulation: Detecting overlapping intervals in a collection of ranges can be crucial for scheduling, resource allocation, and data analysis tasks. We’ll consider ‘intervals’ as pairs of numbers, and an overlap occurs when two intervals share at least one number. The goal is to identify these overlaps and return the resulting intervals sorted in ascending order. For example, given input intervals [(1, 3), (2, 6), (8, 10)], the desired output is [(1, 6), (8, 10)].

Method 1: Brute Force Approach

The brute force method involves comparing each interval with all other intervals to check for overlaps. It’s straightforward but not time-efficient, especially for large datasets.

Here’s an example:

def find_overlapping_intervals(intervals):
    intervals.sort(key=lambda x: x[0])  # Sort by start times
    merged = [intervals[0]]
    for current in intervals[1:]:
        last = merged[-1]
        if current[0] <= last[1]:  # Check for overlap
            merged[-1] = (last[0], max(last[1], current[1]))  # Merge overlapping intervals
        else:
            merged.append(current)
    return merged

# Example intervals
example_intervals = [(1, 3), (2, 6), (8, 10)]
print(find_overlapping_intervals(example_intervals))

Output:

[(1, 6), (8, 10)]

This code snippet starts by sorting the interval list by each interval’s starting point. Then, it initialises a new list with the first interval and iterates through the rest, checking for overlaps and merging them as necessary. Non-overlapping intervals are appended as they are.

Method 2: Using the IntervalTree library

The IntervalTree library provides a data structure that is particularly useful for interval queries. This method is efficient for large sets of intervals and involves less coding.

Here’s an example:

from intervaltree import IntervalTree

def find_overlapping_intervals(intervals):
    tree = IntervalTree()
    for interval in intervals:
        tree[interval[0]:interval[1]] = True
    tree.merge_overlaps()  # Merge overlapping intervals
    return sorted((int(begin), int(end)) for begin, end, _ in tree)

# Example intervals
example_intervals = [(1, 3), (2, 6), (8, 10)]
print(find_overlapping_intervals(example_intervals))

Output:

[(1, 6), (8, 10)]

In this example, IntervalTree from the intervaltree library is used to merge overlapping intervals efficiently. After adding all the intervals to the tree, the merge_overlaps() method is called, which combines all overlapping intervals. The result is then sorted and returned.

Method 3: Sweep Line Technique

The sweep line technique involves ‘sweeping’ a line across the x-axis and recording the start and end of each interval. It’s an elegant and often fast technique for detecting and merging overlapping intervals.

Here’s an example:

def find_overlapping_intervals(intervals):
    events = sorted([(start, 1) for start, end in intervals] + [(end, -1) for start, end in intervals])
    merged, count = [], 0
    for point, event in events:
        if event == 1:  # Interval starts
            if count == 0:
                merged.append([point, None])
            count += 1
        else:  # Interval ends
            if count == 1:
                merged[-1][1] = point
            count -= 1
    return [tuple(interval) for interval in merged]

# Example intervals
example_intervals = [(1, 3), (2, 6), (8, 10)]
print(find_overlapping_intervals(example_intervals))

Output:

[(1, 6), (8, 10)]

This code uses the sweep line technique which is powerful for managing interval problems efficiently. It sorts the events (start and end points) and sweeps through them, counting active intervals and merging as needed. The result is a list of merged intervals.

Method 4: Sorting and Merging

This method improves upon the brute force approach by carefully sorting interval starting points and then merging overlapping ranges in a single pass, which can be more efficient than a naive pairwise comparison.

Here’s an example:

def find_overlapping_intervals(intervals):
    if not intervals:
        return []
    # Sort by starting points of the intervals
    intervals.sort(key=lambda x: x[0])
    merged = [intervals[0]]
    for i in range(1, len(intervals)):
        # If the intervals overlap, merge them
        if merged[-1][1] >= intervals[i][0]:
            merged[-1] = (merged[-1][0], max(merged[-1][1], intervals[i][1]))
        else:
            merged.append(intervals[i])
    return merged

# Example intervals
example_intervals = [(1, 3), (2, 6), (8, 10)]
print(find_overlapping_intervals(example_intervals))

Output:

[(1, 6), (8, 10)]

This snippet is quite similar to the first method but focuses on improving clarity and performance through single-pass merging. Sorting starts first to ensure that overlapping intervals are adjacent, then those intervals are merged sequentially.

Bonus One-Liner Method 5: Using heapq

Utilizing the heapq library in Python, this one-liner cleverly merges intervals in sorted order using a heap data structure for maintaining the minimal end-point at the top.

Here’s an example:

import heapq

def find_overlapping_intervals(intervals):
    if not intervals:
        return []
    intervals.sort()
    heap = [intervals[0][1]]
    for start, end in intervals[1:]:
        if start > heap[0]:
            heapq.heappop(heap)
        heapq.heappush(heap, end)
    return intervals

# Example intervals
example_intervals = [(1, 3), (2, 6), (8, 10)]
print(find_overlapping_intervals(example_intervals))

Output:

[(1, 3), (2, 6), (8, 10)]

This method is concise but somewhat misleading as it doesn’t return the merged intervals by default; it’s more of a foundation for an approach that would need to be extended to handle the merging. It sorts the intervals and uses a heap to track the ending times.

Summary/Discussion

Method 1:

Method 2:

Method 3:

Method 4:

Method 5: