intersection-of-two-arrays.py

# Time:  O(m + n)
# Space: O(min(m, n))

# Given two arrays, write a function to compute their intersection.
#
# Example:
# Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].
#
# Note:
# Each element in the result must be unique.
# The result can be in any order.

# Hash solution.


class Solution(object):
    def intersection(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        if len(nums1) > len(nums2):
            return self.intersection(nums2, nums1)
  
        lookup = set()
        for i in nums1:
            lookup.add(i)

        res = []
        for i in nums2:
            if i in lookup:
                res += i,
                lookup.discard(i)

        return res

    def intersection2(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        return list(set(nums1) & set(nums2))


# Time:  O(max(m, n) * log(max(m, n)))
# Space: O(1)
# Binary search solution.
class Solution2(object):
    def intersection(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        if len(nums1) > len(nums2):
            return self.intersection(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()

        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 = binary_search(lambda x, y: x > y, nums2, left, len(nums2), i)
        
        return res


# Time:  O(max(m, n) * log(max(m, n)))
# Space: O(1)
# Two pointers solution.
class Solution3(object):
    def intersection(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        nums1.sort(), nums2.sort()
        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:
                if not res or res[-1] != nums1[it1]:
                    res += nums1[it1],
                it1 += 1
                it2 += 1
        
        return res