IntersectionOfTwoArraysII

Problem

https://leetcode.com/problems/intersection-of-two-arrays-ii/

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?

Solution

Count the number of times each number appears in each array in 2 dictionaries dict1 and dict2. For each key in dict1 and dict2, add key repeated

min(dict1[key], dict2[key])

times to the intersection list.

Pattern

Array, Hash Table, Two Pointers, Binary Search, Sorting

Code

from typing import List


def intersect(nums1: List[int], nums2: List[int]) -> List[int]:
    dict1 = counts(nums1)
    dict2 = counts(nums2)

    intersection = []
    for key, val in dict1.items():
        if key in dict2:
            repeat = min(val, dict2[key])
            intersection.extend(repeat * [key])

    return intersection


def counts(nums):
    d = {}
    for n in nums:
        if n in d:
            d[n] += 1
        else:
            d[n] = 1
    return d

Test

>>> from IntersectionOfTwoArraysII import intersect
>>> intersect([1,2,2,1], [2,2])
[2, 2]
>>> intersect([4,9,5], [9,4,9,8,4])
[4, 9]

IntersectionOfTwoArraysII.counts(nums)

IntersectionOfTwoArraysII.intersect(nums1: List[int], nums2: List[int]) → List[int]