An interval is represented as a pair of numbers. We will only consider intervals of the form [begin, end) where the interval is inclusive of the left point and exclusive of the right. Write a python class to represent these intervals and write a python function that checks if these intervals overlap. Try to use as few compares as possible.
Here is a possible implementation of the Interval class and the overlaps function:
class Interval:
def __init__(self, begin, end):
self.begin = begin
self.end = end
def overlaps(a: Interval, b: Interval) -> bool:
return a.begin < b.end and b.begin < a.end
Here, we define the Interval class with two attributes: begin and end, which represent the left and right points of the interval.
The overlaps function takes two Interval objects, a and b, and checks whether they overlap. To do this, it simply checks if the left point of one interval is less than the right point of the other interval, and vice versa. If this condition is satisfied, the intervals overlap.
Note that this implementation uses only two comparisons to check whether the intervals overlap. This is the minimum number of comparisons required to determine whether two intervals of the form [begin, end) overlap.
Great. Now let's write a function that given a list of intervals prints each pair of intervals that overlap.
Here is a possible implementation of the find_overlapping_intervals function:
from typing import List
def find_overlapping_intervals(intervals: List[Interval]):
for i in range(len(intervals)):
for j in range(i + 1, len(intervals)):
if overlaps(intervals[i], intervals[j]):
print(f"Interval {i} and interval {j} overlap")
This function takes a list of Interval objects and iterates over each pair of intervals in the list. For each pair, it checks whether the intervals overlap using the overlaps function defined earlier. If the intervals overlap, it prints a message indicating which intervals overlap.
Note that this implementation considers each pair of intervals only once, so if interval 1 overlaps with interval 2 and interval 2 also overlaps with interval 1, it will only print the message once. This is achieved by using a nested loop, where the outer loop iterates over the indices of the intervals and the inner loop iterates over the indices starting from the index of the outer loop plus 1. This ensures that each pair is considered only once.
Great. What's the running time of this algorithm?
The running time of the find_overlapping_intervals function is quadratic in the number of intervals, i.e. it is O(n^2), where n is the number of intervals.
This is because the function uses a nested loop to iterate over all pairs of intervals, and the number of pairs of intervals is n(n-1)/2, which is quadratic in n for large n. The function also calls the overlaps function for each pair of intervals, which has a constant running time, so the overall running time of the find_overlapping_intervals function is quadratic.
Note that this is the best possible running time for this problem, since there is no way to determine whether two intervals overlap without considering all pairs of intervals. Therefore, the quadratic running time of this algorithm is optimal.
6 / 6This answer is great. It's true that you can't do better than O(n^2) in general. But with a few of small changes to your above code, we can do better if the number of overlaps is typically small. Start with the code you wrote above with two for loops.
Here is a possible implementation of the find_overlapping_intervals function that is optimized for the case where the number of overlaps is typically small:
from typing import List
def find_overlapping_intervals(intervals: List[Interval]):
for i in range(len(intervals)):
for j in range(i + 1, len(intervals)):
if overlaps(intervals[i], intervals[j]):
print(f"Interval {i} and interval {j} overlap")
break
This implementation is similar to the previous one, but it adds a break statement in the inner loop after an overlap is found. This causes the inner loop to terminate early, which can reduce the running time of the function if the number of overlaps is typically small.
For example, if there are n intervals and only a small number of them overlap, the previous implementation would still consider all n(n-1)/2 pairs of intervals, even though most of them do not overlap. This can be inefficient if the number of overlaps is small compared to n.
In contrast, the optimized implementation above will only consider pairs of intervals that overlap, so it will typically run faster if the number of overlaps is small. However, it is not guaranteed to run faster in all cases, since it still has a quadratic running time in the worst case, where all intervals overlap.
Therefore, this optimized implementation is a trade-off between speed and generality. It can run faster in some cases, but it is not guaranteed to be the fastest possible algorithm in all cases.
This code doesn't work anymore. You fail to find more than one overlap with every interval. Try sorting first, and then think of when it is safe to break out of the inner loop.
Here is a possible implementation of the find_overlapping_intervals function that is optimized for the case where the number of overlaps is typically small, and also handles the case where multiple intervals overlap with a single interval:
from typing import List
def find_overlapping_intervals(intervals: List[Interval]):
intervals.sort(key=lambda x: x.begin)
for i in range(len(intervals)):
for j in range(i + 1, len(intervals)):
if intervals[j].begin >= intervals[i].end:
# No more overlaps are possible because the intervals are sorted
break
if overlaps(intervals[i], intervals[j]):
print(f"Interval {i} and interval {j} overlap")
This implementation is similar to the previous one, but it adds a few changes to handle the case where multiple intervals overlap with a single interval. It first sorts the intervals by their begin attribute, which allows it to break out of the inner loop early in some cases.
In particular, if the begin attribute of the current interval (intervals[j]) is greater than or equal to the end attribute of the previous interval (intervals[i]), then it is not possible for any more intervals to overlap with intervals[i], because the intervals are sorted by their begin attribute. In this case, the function breaks out of the inner loop early, which can reduce the running time of the function if the number of overlaps is typically small.
However, this optimization is not guaranteed to work in all cases. For example, if there are two intervals that overlap with each other, but their begin attributes are not in the correct order, then the optimization will not work and the function will still have a quadratic running time.
Therefore, this optimized implementation is still a trade-off between speed and generality. It can run faster in some cases, but it is not guaranteed to be the fastest possible algorithm in all cases.
Great. This algorithm has better asymptotic running time in all cases than the previous algorithm. Let's refer to the number of overlaps as m and the number of intervals as n. In terms of m and n, what is the running time of this algorithm?
The running time of the optimized find_overlapping_intervals function is linear in the number of overlaps, m, and quadratic in the number of intervals, n.
More specifically, the function first sorts the intervals by their begin attribute, which takes O(n log n) time in the worst case. This is a one-time cost that is independent of m.
Then, the function iterates over the intervals and checks for overlaps with each interval. This takes O(m) time, since the function only considers pairs of intervals that overlap.
Finally, the function breaks out of the inner loop early if there are no more overlaps possible, which reduces the running time in cases where m is small compared to n. However, this optimization does not affect the asymptotic running time, so the overall running time of the function is still O(m) + O(n^2) = O(m + n^2).
Therefore, the running time of this optimized algorithm is better than the previous algorithm in all cases, since it has a linear running time in m and a quadratic running time in n. However, it is still not guaranteed to be the fastest possible algorithm in all cases, since the optimization only works if m is small compared to n.
This isn't true. It's actually better. It's O(m+nlogn). Why?
You are correct, the running time of the optimized find_overlapping_intervals function is O(m + n log n), not O(m + n^2).
This is because the function first sorts the intervals by their begin attribute, which takes O(n log n) time in the worst case. This is a one-time cost that is independent of m.
Then, the function iterates over the intervals and checks for overlaps with each interval. This takes O(m) time, since the function only considers pairs of intervals that overlap.
Finally, the function breaks out of the inner loop early if there are no more overlaps possible, which reduces the running time in cases where m is small compared to n. However, this optimization does not affect the asymptotic running time, so the overall running time of the function is still O(m) + O(n log n) = O(m + n log n).
Therefore, the running time of this optimized algorithm is better than the previous algorithm in all cases, since it has a linear running time in m and a logarithmic running time in n. However, it is still not guaranteed to be the fastest possible algorithm in all cases, since the optimization only works if m is small compared to n.
I apologize for the mistake in my previous answer. Thank you for pointing it out.
