Outpost must so possible.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
Related questions
Question
Outpost must so possible.
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"""Given a list of points, find the k closest to the origin.
Idea: Maintain a max heap of k elements.
We can iterate through all points.
If a point p has a smaller distance to the origin than the top element of a
heap, we add point p to the heap and remove the top element.
After iterating through all points, our heap contains the k closest points to
the origin.
111111
from heapq import heapify, heappushpop
def k_closest(points, k, origin=(0, 0)):
# Time: 0(k+(n-k)logk)
# Space: 0(k)
"""Initialize max heap with first k points.
Python does not support a max heap; thus we can use the default min heap
where the keys (distance) are negated.
heap = [(-distance (p, origin), p) for p in points[:k]]
heapify(heap)
11 11 11
For every point p in points [k:],
check if p is smaller than the root of the max heap;
if it is, add p to heap and remove root. Reheapify.
11 11 11
for point in points [k:]:
dist = distance (point, origin)
heappushpop (heap, (-dist, point)) #heappushpop does conditional check
"""Same as:
if d < -heap [0][0]:
heappush(heap, (-d,p))
heappop (heap)
Note: heappushpop is more efficient than separate push and pop calls.
Each heappushpop call takes 0(logk) time.
return [point for nd, point in heap] # return points in heap
def distance (point, origin=(0, 0)):
""" Calculates the distance for a point from origo"""
return (point [0] - origin[0])**2 + (point [1] - origin[1]) **2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a48f55f-d319-4fa0-a1c3-0d0c662e6918%2Faed67b2d-9ce4-46ed-9f49-b59c1042ee06%2F5l0sdac_processed.jpeg&w=3840&q=75)
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"""Given a list of points, find the k closest to the origin.
Idea: Maintain a max heap of k elements.
We can iterate through all points.
If a point p has a smaller distance to the origin than the top element of a
heap, we add point p to the heap and remove the top element.
After iterating through all points, our heap contains the k closest points to
the origin.
111111
from heapq import heapify, heappushpop
def k_closest(points, k, origin=(0, 0)):
# Time: 0(k+(n-k)logk)
# Space: 0(k)
"""Initialize max heap with first k points.
Python does not support a max heap; thus we can use the default min heap
where the keys (distance) are negated.
heap = [(-distance (p, origin), p) for p in points[:k]]
heapify(heap)
11 11 11
For every point p in points [k:],
check if p is smaller than the root of the max heap;
if it is, add p to heap and remove root. Reheapify.
11 11 11
for point in points [k:]:
dist = distance (point, origin)
heappushpop (heap, (-dist, point)) #heappushpop does conditional check
"""Same as:
if d < -heap [0][0]:
heappush(heap, (-d,p))
heappop (heap)
Note: heappushpop is more efficient than separate push and pop calls.
Each heappushpop call takes 0(logk) time.
return [point for nd, point in heap] # return points in heap
def distance (point, origin=(0, 0)):
""" Calculates the distance for a point from origo"""
return (point [0] - origin[0])**2 + (point [1] - origin[1]) **2
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