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intersection-of-two-arrays-ii.py
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# If the given array is not sorted and the memory is unlimited:
# - Time: O(m + n)
# - Space: O(min(m, n))
# elif the given array is already sorted:
# if m << n or m >> n:
# - Time: O(min(m, n) * log(max(m, n)))
# - Space: O(1)
# else:
# - Time: O(m + n)
# - Soace: O(1)
# else: (the given array is not sorted and the memory is limited)
# - Time: O(max(m, n) * log(max(m, n)))
# - Space: O(1)
# Given two arrays, write a function to compute their intersection.
#
# Example:
# Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].
#
# Note:
# Each element in the result should appear as many times as it shows in both arrays.
# The result can be in any order.
#
# Follow up:
# - What if the given array is already sorted? How would you optimize your algorithm?
# - What if nums1's size is small compared to num2's size? Which algorithm is better?
# - What if elements of nums2 are stored on disk, and the memory is limited such that
# you cannot load all elements into the memory at once?
# If the given array is not sorted and the memory is unlimited.
# Time: O(m + n)
# Space: O(min(m, n))
# Hash solution.
class Solution(object):
def intersect(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
if len(nums1) > len(nums2):
return self.intersect(nums2, nums1)
lookup = collections.defaultdict(int)
for i in nums1:
lookup[i] += 1
res = []
for i in nums2:
if lookup[i] > 0:
res += i,
lookup[i] -= 1
return res
# If the given array is already sorted, and the memory is limited, and (m << n or m >> n).
# Time: O(min(m, n) * log(max(m, n)))
# Space: O(1)
# Binary search solution.
class Solution(object):
def intersect(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
if len(nums1) > len(nums2):
return self.intersect(nums2, nums1)
def binary_search(compare, nums, left, right, target):
while left < right:
mid = left + (right - left) / 2
if compare(nums[mid], target):
right = mid
else:
left = mid + 1
return left
nums1.sort(), nums2.sort() # Make sure it is sorted, doesn't count in time.
res = []
left = 0
for i in nums1:
left = binary_search(lambda x, y: x >= y, nums2, left, len(nums2), i)
if left != len(nums2) and nums2[left] == i:
res += i,
left += 1
return res
# If the given array is already sorted, and the memory is limited or m ~ n.
# Time: O(m + n)
# Soace: O(1)
# Two pointers solution.
class Solution(object):
def intersect(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
nums1.sort(), nums2.sort() # Make sure it is sorted, doesn't count in time.
res = []
it1, it2 = 0, 0
while it1 < len(nums1) and it2 < len(nums2):
if nums1[it1] < nums2[it2]:
it1 += 1
elif nums1[it1] > nums2[it2]:
it2 += 1
else:
res += nums1[it1],
it1 += 1
it2 += 1
return res
# If the given array is not sorted, and the memory is limited.
# Time: O(max(m, n) * log(max(m, n)))
# Space: O(1)
# Two pointers solution.
class Solution(object):
def intersect(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
nums1.sort(), nums2.sort() # O(max(m, n) * log(max(m, n)))
res = []
it1, it2 = 0, 0
while it1 < len(nums1) and it2 < len(nums2):
if nums1[it1] < nums2[it2]:
it1 += 1
elif nums1[it1] > nums2[it2]:
it2 += 1
else:
res += nums1[it1],
it1 += 1
it2 += 1
return res