intersection-of-two-arrays-ii.py

"""

350. Intersection of Two Arrays II
Easy

Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.

 
Example 1:

Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]
Example 2:

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]
Explanation: [9,4] is also accepted.
 

Constraints:

1 <= nums1.length, nums2.length <= 1000
0 <= nums1[i], nums2[i] <= 1000
 

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 nums2'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?

"""

# V0
# IDEA : 2 POINTERS
class Solution(object):
    def intersect(self, nums1, nums2):
        res = []
        nums1.sort()
        nums2.sort()
        i = j = 0
        while i < len(nums1) and j < len(nums2):
            if nums1[i] == nums2[j]:
                res.append(nums1[i])
                i += 1
                j += 1
            elif nums1[i] < nums2[j]:
                i += 1
            elif nums1[i] > nums2[j]:
                j += 1
        return res
  
# V0' 
class Solution(object):
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        return list((collections.Counter(nums1) & collections.Counter(nums2)).elements())

# V1 
# https://blog.csdn.net/coder_orz/article/details/51496278
# IDEA : COUNTER
class Solution(object):
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        return list((collections.Counter(nums1) & collections.Counter(nums2)).elements())

# V1' 
# https://blog.csdn.net/coder_orz/article/details/51496278
# IDEA : TWO POINTER 
class Solution(object):
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        res = []
        nums1.sort()
        nums2.sort()
        i = j = 0
        while i < len(nums1) and j < len(nums2):
            if nums1[i] == nums2[j]:
                res.append(nums1[i])
                i += 1
                j += 1
            elif nums1[i] < nums2[j]:
                i += 1
            elif nums1[i] > nums2[j]:
                j += 1

        return res

# V1'' 
# https://blog.csdn.net/coder_orz/article/details/51496278
# IDEA : DICT 
class Solution(object):
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        res = []
        map = {}
        for i in nums1:
            map[i] = map[i]+1 if i in map else 1
        for j in nums2:
            if j in map and map[j] > 0:
                res.append(j)
                map[j] -= 1

        return res

# V2 
# Time:  O(m + n)
# Space: O(min(m, n))
# Hash solution.
import collections
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

    def intersect2(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        c = collections.Counter(nums1) & collections.Counter(nums2)
        intersect = []
        for i in c:
            intersect.extend([i] * c[i])
        return intersect

# V3 
# 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

# V4 
# 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

# V5 
# 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