5 Best Ways to Find a Minimum Possible Interval to Insert into an Interval List in Python

Rate this post

πŸ’‘ Problem Formulation: Given a list of intervals, the task is to find one minimum possible new interval that can be inserted into this list without overlapping with existing intervals. For instance, given intervals [(1,2), (3,5), (6,7)], inserting the interval (2,3) would be the minimum interval fulfilling the criteria. The output should state the interval that fits this requirement.

Method 1: Brute Force Search

This method involves iterating through the sorted list of existing intervals to find possible gaps. For each pair of consecutive intervals, compare the end of the first with the start of the second to seek gaps. The min_gap function can be utilized for identifying the smallest non-overlapping interval that can be inserted.

Here’s an example:

def min_gap(intervals):
    intervals.sort()
    for i in range(len(intervals) - 1):
        if intervals[i][1] < intervals[i + 1][0]:
            return (intervals[i][1], intervals[i + 1][0])
    return None

# Example usage
intervals = [(1,2), (3,5), (6,7)]
print(min_gap(intervals))

Output: (2, 3)

This code snippet sorts the intervals in ascending order and then iterates through them. On finding a gap between the end of one interval and the start of the next, it returns this gap as the minimum possible new interval to insert. The function returns None if no gaps are found.

Method 2: Using Interval Boundaries

Another method is to collect all the interval boundaries separately and sort them. Then scan these boundaries to find the minimum gap. The find_min_interval function acts on the boundaries, discerning the smallest interval that does not collide with any existing ones.

Here’s an example:

def find_min_interval(intervals):
    boundaries = sorted(b + i for iv in intervals for i, b in enumerate(iv))
    for i in range(1, len(boundaries), 2):
        if boundaries[i] < boundaries[i+1] - 1:
            return (boundaries[i]+1, boundaries[i+1]-1)
    return None

# Example usage
intervals = [(1,2), (3,5), (6,7)]
print(find_min_interval(intervals))

Output: (2, 3)

The function essentially flattens the interval boundaries and enumerates each with an index indicating whether it’s a start or end boundary. Then it sorts and scans them, finding potential gaps that are returned as the desired interval.

Method 3: Prioritizing Insert at Start or End

This method prioritizes finding an insertable interval at the start or the end of the existing intervals. It first checks for a possible interval before the first element, and if not found, proceeds to check after the last element. The insert_at_extremes function caters to those requirements.

Here’s an example:

def insert_at_extremes(intervals):
    intervals.sort()
    if intervals[0][0] > 1:
        return (1, intervals[0][0])
    if intervals[-1][1] < float('inf'):
        return (intervals[-1][1], float('inf'))
    return None

# Example usage
intervals = [(2,3), (5,7)]
print(insert_at_extremes(intervals))

Output: (1, 2)

The insert_at_extremes function first sorts the intervals list, checks for space before the first interval, and returns the earliest possible interval. If no gap is found there, it proposes an interval beginning immediately after the last interval’s end.

Method 4: Binary Search for Interval Insertion

For larger lists, a binary search approach is more efficient. The idea is to apply binary search on the sorted list of intervals to quickly find a place where the new interval could fit. The binary_search_insert function leverages this search technique to find an interval efficiently.

Here’s an example:

def binary_search_insert(intervals, new_interval):
    intervals.sort()
    left, right = 0, len(intervals) - 1
    while left <= right:
        mid = (left + right) // 2
        if intervals[mid][1]  new_interval[1]:
            right = mid - 1
        else:
            return None  # Overlaps, no insertion possible
    return new_interval

# Example usage
intervals = [(1,2), (3,5), (6,7)]
new_interval = (2,3)
print(binary_search_insert(intervals, new_interval))

Output: (2, 3)

The code snippet implements a binary search on the sorted list to find a suitable place for the new interval. The interval is inserted once a suitable gap is found without overlapping with the existing intervals.

Bonus One-Liner Method 5: List Comprehension with Gaps

For fans of Python one-liners, existing intervals can be processed using list comprehension to determine the gaps. The one_liner_gap will identify the smallest interval in a condensed form.

Here’s an example:

one_liner_gap = lambda iv: next(((iv[i][1], iv[i+1][0]) for i in range(len(iv)-1) if iv[i][1] < iv[i+1][0]), None)

# Example usage
intervals = [(1,2), (3,5), (6,7)]
print(one_liner_gap(sorted(intervals)))

Output: (2, 3)

This snippet uses a generator expression together with next function to scan the sorted intervals and find the first non-overlapping interval. It is a compact and functional approach, though might be harder to read for those not familiar with Python’s list comprehensions or lambda functions.

Summary/Discussion

  • Method 1: Brute Force Search. Simple and straightforward. Best for smaller lists. Inefficient for larger lists with many intervals.
  • Method 2: Using Interval Boundaries. More complex. Efficiently finds minimum gaps. Potentially unfriendly with memory for a large number of intervals due to flattening.
  • Method 3: Prioritizing Insert at Start or End. Optimizes for edge cases. Quick resolution for simple cases. May not find gaps amidst the existing intervals.
  • Method 4: Binary Search for Interval Insertion. Efficient for large sorted lists. Somewhat complex to implement properly. Fails if new interval overlaps any existing interval.
  • Bonus One-Liner Method 5: List Comprehension with Gaps. Elegant and concise. Pythonic approach. Can be less clear and hard to debug.