Problem Description

In this problem, we are provided with two arrays nums1 and nums2, each containing a set of integers. These arrays are 0-indexed, meaning that their indices start from 0. The main objective is to determine two specific values:

1. The first value is the number of elements in nums1 that are also present in nums2. This doesn't mean exact index matching; as long as a value from nums1 shows up anywhere in nums2, it counts.

2. The second value is similar but in the opposite direction: we count the number of elements in nums2 that are also present in nums1.

Our goal is to return these two counts as an array containing two integers, with the first value derived from nums1 and the second from nums2.

Intuition

For this problem, the intuitive approach is to effectively and efficiently determine whether elements of one array occur within the other. We can simplify the task by converting the arrays into sets, which allow us to check for the presence of an element in constant time (ignoring the initial cost of creating the set). This is due to the fact that sets in Python are implemented as hash tables, providing an average-case time complexity of O(1) for querying the presence of an item.

We follow these steps:

1. Convert nums1 and nums2 into sets s1 and s2. This not only provides us with quick lookups but also removes duplicates, which are irrelevant to our counts since we're only interested in whether an element from one array is present at least once in the other.

2. For the first count, we iterate through each element in nums1 (not s1, as we want the original array's indices) and check if it exists in s2. We sum these checks to get the total.

3. For the second count, we perform a similar iteration, but now we loop over nums2 and check for the presence of each element in s1. We again sum the checks to get the total.

4. These two sums give us the required values, and we return them as an array of size 2.

This approach gives us a solution that is efficient and takes full advantage of Python's set structure to minimize the time spent on lookups.

Solution Approach

The implementation of the solution uses the hash table data structure in Python, which is represented by the set. Sets are particularly useful in this scenario because they automatically handle duplicates and allow for efficient membership tests, which is exactly what we need.

Here is a walkthrough of the algorithm:

1. Convert Lists to Sets:

   • We begin by converting the input lists nums1 and nums2 into sets, named s1 and s2. Converting to sets eliminates duplicates and enables us to perform quick lookups.
   • This is done using the syntax s1, s2 = set(nums1), set(nums2).

2. Count Elements in Intersection: Since we are tasked with counting the elements of nums1 that are present in nums2 and vice versa, we proceed as follows:

   • For the first value, we iterate over each element in the original list nums1 using list comprehension and check if the element is in s2 (the set version of nums2). We use the expression sum(x in s2 for x in nums1) to count the number of True outcomes (which happen when x is in s2), effectively giving us the count of intersection elements pertaining to nums1.
   • Similarly, for the second value, we iterate over nums2 and check each element's presence in s1. Using the expression sum(x in s1 for x in nums2), we get the count of intersection elements pertaining to nums2.

3. Return the Result:

   • After calculating both counts, we combine them into a list [sum(x in s2 for x in nums1), sum(x in s1 for x in nums2)] and return this list as our final result.

In terms of complexity:

• Creating the sets from the lists has a time complexity of O(n) for nums1 and O(m) for nums2, where n and m are the lengths of the respective lists.
• The list comprehensions that count the presences will also have a time complexity of O(n) and O(m) for nums1 and nums2 respectively.
• Thus, the overall time complexity of this solution is O(n + m), which is quite efficient given that the lookups in the sets are O(1) operations on average.

By using sets for membership checks and list comprehensions for the counts, we achieve a concise and efficient implementation that aligns closely with the problem's requirements.

Example Walkthrough

Let's illustrate the solution approach with a small example. Suppose we have two arrays:

• nums1 = [1, 2, 2, 3]
• nums2 = [2, 3, 4, 2]

We want to find the number of elements in nums1 that are present in nums2 and vice versa.

First, we convert nums1 and nums2 to sets to remove duplicates and allow efficient lookups. After conversion:

• s1 = {1, 2, 3}
• s2 = {2, 3, 4}

Next, we count the elements in nums1 that are also in s2:

• Element 1 in nums1 is not in s2, count = 0
• Element 2 in nums1 is in s2, count = 1
• Element 2 in nums1 is in s2, count = 2 (repeated element, but we still count it)
• Element 3 in nums1 is in s2, count = 3

So, there are 3 elements in nums1 that are present in nums2. We note that even though '2' is a duplicate in nums1, it is counted twice as it appears twice in the original array.

We repeat the process for nums2 against s1:

• Element 2 in nums2 is in s1, count = 1
• Element 3 in nums2 is in s1, count = 2
• Element 4 in nums2 is not in s1, count = 2
• Element 2 in nums2 is in s1, count = 3 (again, we count the duplicate)

We find there are 3 elements in nums2 that are present in nums1.

Lastly, we return the result as an array: [3, 3], representing the counts as required.

Through this example, we see how the solution approach handles array elements and their presence in the other array by using set operations and iteration to count the number of occurrences efficiently.

Solution Implementation

Time and Space Complexity

The given code determines the number of intersection values between two lists, nums1 and nums2. The computational complexity is as follows:

Time Complexity

The time complexity of the code is O(n + m), where n is the length of nums1 and m is the length of nums2.

Initially, we convert both lists nums1 and nums2 into sets, which takes O(n) and O(m) time respectively. Then we compute the sum of the boolean expressions checking if each element of nums1 is in s2 and if each element of nums2 is in s1. Both of these operations are done in O(n) and O(m) respectively, since set lookup is O(1) on average. Hence, in total, the time complexity is O(n) + O(m) for both conversions and the sum operations.

Space Complexity

The space complexity of the code is O(n + m) as well. This is because it creates two additional sets from the lists nums1 and nums2, holding up to n and m unique elements respectively, assuming the worst case where all elements in the lists are unique. There is no other significant use of additional space.

Learn more about how to find time and space complexity quickly using problem constraints.