Problem Description

The problem gives us three integer arrays named nums1, nums2, and nums3. Our task is to find a distinct array that contains all the unique values which are present in at least two of these arrays. This means if a value is present in either array nums1 & nums2, nums1 & nums3, or nums2 & nums3, it should be included in the final answer. Also, the result should not have any duplicates, and the order of values in the resulting array does not matter.

Intuition

To solve this problem efficiently, we can use the properties of set data structures in Python, which store only unique elements. First, we convert each input array (nums1, nums2, nums3) to a set to eliminate any duplicates within the individual arrays. This step helps to simplify our upcoming operations since sets do not allow duplicate values, and membership checks (to see if a value is present in the set) are done in constant time.

The next step is to iterate through a range of possible values that could be present in the arrays. Since the problem does not specify an upper limit on the values, we can assume that the numbers are within a reasonable domain. In this solution, it is assumed that the values are within the inclusive range of 1 to 100.

For each number in this range, we check if it is present in at least two of the three sets. To do this, we find the sum of the boolean results of the membership tests for the three sets (i.e., (i in s1) + (i in s2) + (i in s3)). If a number is present in at least two sets, this sum will be greater than 1. We use a list comprehension to build the final list of numbers that meet this criterion.

This approach is concise and leverages the strengths of Python's data structures to arrive at a solution that is not only efficient but also easy to understand.

Solution Approach

The implementation of the solution is straightforward yet efficient, leveraging the power of Python's set data structures and list comprehensions. Here's a step-by-step breakdown of the algorithm:

1. Convert Lists to Sets: Since we are only interested in the presence of values rather than their frequency, the first step is to convert the given lists nums1, nums2, and nums3 into sets. This is done using the Python built-in set constructor. By doing so, we remove any duplicates within each of the individual lists, and it also allows us to perform the next steps more efficiently.

   s1, s2, s3 = set(nums1), set(nums2), set(nums3)

2. Iterate Over a Range of Values: Given the problem does not provide a specific range for the integer values, the solution assumes a reasonable range from 1 to 100, inclusive. Since there's no information that any value outside this range will appear in the arrays, it is safe to iterate over these 100 numbers.

   range(1, 101)

3. Check for Presence in Two Sets: For each value i in the range, we check if i is present in at least two out of the three sets (s1, s2, s3). This is done using a simple sum of boolean expressions:

   (i in s1) + (i in s2) + (i in s3) > 1

   Each expression (i in s1), (i in s2), or (i in s3) returns a Boolean value—True if i is a member of the set and False otherwise. When these are added together, True counts as 1 and False as 0. If the sum is greater than 1, it indicates that i is present in at least two sets.

4. List Comprehension to Create the Result: A list comprehension is used to iterate through the range and apply the check. Only values that pass the check (present in at least two sets) are included in the final list. The beauty of using a list comprehension here is that it returns a new list directly, without having to manually initialize an empty list and append qualifying values one by one.

   [i for i in range(1, 101) if (i in s1) + (i in s2) + (i in s3) > 1]

By incorporating these steps into the provided twoOutOfThree method of the Solution class, we get a compact yet elegant solution that meets the problem's requirements effectively. This solution approach demonstrates the strength of using sets for membership tests, the efficiency of list comprehensions for building lists, and some basic mathematical operations (like summing boolean values) for conditional checks.

Example Walkthrough

Let's examine the solution approach with a small example. Assume the input arrays are as follows:

nums1 = [1, 2, 3]
nums2 = [2, 4]
nums3 = [3, 4, 5]

Applying the solution approach step-by-step:

1. Convert Lists to Sets: We convert nums1, nums2, and nums3 to sets to remove any duplicates and to simplify the upcoming membership checks:

   s1 = set(nums1) # s1 = {1, 2, 3}
   s2 = set(nums2) # s2 = {2, 4}
   s3 = set(nums3) # s3 = {3, 4, 5}

2. Iterate Over a Range of Values: We loop through the range from 1 to 100, as we are assuming a reasonable domain for the numbers:

   range(1, 101) # Will iterate from 1 to 100 inclusive

3. Check for Presence in Two Sets: For each number i from 1 to 100, we perform the membership tests to see if i is present in at least two of the sets s1, s2, and s3. For example:

   (2 in s1) + (2 in s2) + (2 in s3) # This equals 2, since 2 is in s1 and s2
   (4 in s1) + (4 in s2) + (4 in s3) # This equals 2, since 4 is in s2 and s3

   Only the numbers meeting the condition > 1 will signify they are present in at least two arrays.

4. List Comprehension to Create the Result: We use a list comprehension to go through the range and collect numbers that meet our condition:

   [i for i in range(1, 101) if (i in s1) + (i in s2) + (i in s3) > 1]

Given our example arrays, the final processing would look like this:

s1, s2, s3 = set(nums1), set(nums2), set(nums3)
result = [i for i in range(1, 101) if (i in s1) + (i in s2) + (i in s3) > 1]

The numbers that are present in at least two of the sets are 2, 3, and 4. Therefore, the result would be:

result = [2, 3, 4]

This small example illustrates how the four steps of the solution approach come together to solve the problem efficiently. Using this method, we quickly find the numbers that are present in at least two of the input arrays and return them as part of the distinct result array.

Solution Implementation

Time and Space Complexity

The given code aims to find all the numbers present in at least two out of the three input lists, and the solution leverages sets for efficient lookups. Here's the analysis of the time and space complexities:

Time Complexity

1. Converting nums1, nums2, and nums3 into sets s1, s2, and s3. This step has a time complexity of O(n) for each list, where n is the length of the longest list among nums1, nums2, and nums3.
2. Iterating through the range [1, 101] and checking the presence of each number in the sets takes O(100) time, since it iterates over a fixed range of numbers.
3. The presence checks i in s1, i in s2, and i in s3 are O(1) operations due to set lookup properties. Since these checks are done for each number in [1, 101], this does not change the overall fixed time iteration.

Considering these steps, the overall time complexity can be approximated to O(n + 100), which simplifies to O(n) because, typically, the constant factors are dropped in complexity analysis.

Space Complexity

1. The additional sets s1, s2, and s3 each have a maximum space complexity of O(n), assuming all elements are unique in the original lists.
2. The final list comprehension doesn't store more than 100 integers since it's constrained by the range [1, 101], which accounts for O(1) space.

Hence, the total space complexity of the code is O(3n + 1), which simplifies to O(n) as constant factors and coefficients are omitted in Big O notation.

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