973. K Closest Points to Origin
Read the full problem statement on LeetCode.
Difficulty: medium Acceptance: 68% Topics: Array, Math, Divide and Conquer, Geometry, Sorting, Heap (Priority Queue), Quickselect
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Reference solution (spoiler · python)
# Time: O(n) on average
# Space: O(1)
# quick select solution
from random import randint
class Solution(object):
def kClosest(self, points, K):
"""
:type points: List[List[int]]
:type K: int
:rtype: List[List[int]]
"""
def dist(point):
return point[0]**2 + point[1]**2
def kthElement(nums, k, compare):
def PartitionAroundPivot(left, right, pivot_idx, nums, compare):
new_pivot_idx = left
nums[pivot_idx], nums[right] = nums[right], nums[pivot_idx]
for i in xrange(left, right):
if compare(nums[i], nums[right]):
nums[i], nums[new_pivot_idx] = nums[new_pivot_idx], nums[i]
new_pivot_idx += 1
nums[right], nums[new_pivot_idx] = nums[new_pivot_idx], nums[right]
return new_pivot_idx
left, right = 0, len(nums) - 1
while left <= right:
pivot_idx = randint(left, right)
new_pivot_idx = PartitionAroundPivot(left, right, pivot_idx, nums, compare)
if new_pivot_idx == k:
return
elif new_pivot_idx > k:
right = new_pivot_idx - 1
else: # new_pivot_idx < k.
left = new_pivot_idx + 1
kthElement(points, K-1, lambda a, b: dist(a) < dist(b))
return points[:K]
# Time: O(nlogk)
# Space: O(k)
import heapq
class Solution2(object):
def kClosest(self, points, K):
"""
:type points: List[List[int]]
:type K: int
:rtype: List[List[int]]
"""
def dist(point):
return point[0]**2 + point[1]**2
max_heap = []
for point in points:
heapq.heappush(max_heap, (-dist(point), point))
if len(max_heap) > K:
heapq.heappop(max_heap)
return [heapq.heappop(max_heap)[1] for _ in xrange(len(max_heap))]
Solution from kamyu104/LeetCode-Solutions · MIT