LC-0349 Easy LeetCode

349. Intersection of Two Arrays

Array Hash Table Two Pointers Binary Search Sorting

Read the full problem statement on LeetCode.

Difficulty: easy Acceptance: 76% Topics: Array, Hash Table, Two Pointers, Binary Search, Sorting

Reading material

Arrays & Hashing 12 min Two Pointers 9 min Binary Search 12 min

Reference solution (spoiler · python)

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

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

Solution from kamyu104/LeetCode-Solutions · MIT