Problem Description

In this problem, we are given an array nums that has an even number of elements. Our goal is to split this array into two halves, nums1 and nums2, each containing exactly half the number of elements of the original array. The key challenge is to ensure that each half contains only distinct elements, meaning no duplicates are allowed within nums1 or nums2.

The problem asks us to determine if such a split is possible and to return true if it is and false otherwise.

To simplify:

• nums has an even number of elements.
• Split nums into two equal halves nums1 and nums2.
• Both nums1 and nums2 must contain only unique elements.
• Decide if the split is achievable or not.

Intuition

The intuition behind the solution lies in the constraint that both halves of the array must contain distinct elements. Since we must split the array into two equally sized parts, each with unique elements, the most immediate issue we would face is if a number appears too many times. Specifically, if a number appears three times or more, it is impossible to split the array into two parts with all distinct elements, as at least one of the parts would end up with duplicates of that number.

By using a counter to tally the frequency of each number, we can easily determine if any number exceeds the limit of two appearances in the array. If the maximum count of any element in the array is less than three, then we can ensure a split where both halves have distinct elements. This leads us to our simple solution approach.

Here's the gist of our solution approach:

1. Count the occurrences of each element in the array using a Counter.
2. Check if any element occurs three or more times.
3. If the maximum count is less than three, a valid split is possible, hence return true.
4. Otherwise, if any element occurs three or more times, return false because the split is impossible.

Solution Approach

The implementation of our solution utilizes the Counter class from Python's collections module, which is a specialized data structure that works like a dictionary. It is designed to count hashable objects, which in this problem are the integers in the nums array. The Counter counts how many times each number appears in the array.

Here's a step-by-step explanation of how the implementation works:

1. Create a Counter object that takes the nums array as input. This will return a Counter object where keys are the distinct numbers from the array and the values are the counts of those numbers.

   1counts = Counter(nums)

2. Use the .values() method of the Counter object to get a list of all the counts (i.e., how many times each number is repeated in the array).

3. Apply the max function on this list to find the highest count. This tells us the maximum number of times any single number appears in nums.

   1max_count = max(counts.values())

4. Finally, if the maximum count is less than 3, which implies no number occurs more than twice, we can make a split with all distinct elements in each part (nums1 and nums2). Hence, the function will return True.

5. If the maximum count is 3 or more, at least one part of the split cannot have all distinct elements. Thus, the function will return False.

Here's the corresponding part of the Python code provided:

1class Solution:
2    def isPossibleToSplit(self, nums: List[int]) -> bool:
3        return max(Counter(nums).values()) < 3

In summary, the solution approach is based on the realization that if any number occurs three times or more, we cannot create two halves with all distinct elements from nums. By using a Counter, we efficiently track the frequency of each element and use this information to determine the possibility of the intended split.

Example Walkthrough

Let's walk through an example to illustrate the solution approach:

Suppose our input array nums is [1, 2, 3, 3, 4, 4]. We want to split this array into two halves, each with three elements and no duplicates within each half.

Here are the steps following the proposed solution approach:

1. We first count the occurrences of each element in nums:

   • The number 1 appears once.
   • The number 2 appears once.
   • The number 3 appears twice.
   • The number 4 appears twice.

   The count looks like this: {1: 1, 2: 1, 3: 2, 4: 2}.

2. Now we check for the highest count among all the elements. In our count, the highest value is 2 as both 3 and 4 appear twice. No number appears three times or more.

3. Since the maximum count is less than 3, it is possible to split the array into two halves with all distinct elements. In this case, one possible split would be:

   • nums1 could be [1, 3, 4]
   • nums2 could be [2, 3, 4]

   Each half has distinct elements, and we have successfully split nums into two valid halves.

To conclude, for the input array [1, 2, 3, 3, 4, 4], our function would return True, indicating that the split is achievable.

Here's how we could translate the example into code using the Counter from Python's collections module:

1from collections import Counter
2
3nums = [1, 2, 3, 3, 4, 4]
4
5# Create a `Counter` object to count the occurrences.
6counts = Counter(nums)
7
8# Get the maximum frequency of any number in `nums`.
9max_count = max(counts.values())
10
11# Determine if a split is possible.
12is_possible_to_split = max_count < 3
13
14print(is_possible_to_split)  # Output: True

For this example, the Python code will print True, as expected from our earlier analysis.

Solution Implementation

Time and Space Complexity

The time complexity of the code can be analyzed based on the operations it performs. The Counter class from the collections module iterates over all elements of nums to count the frequency of each unique number, which requires O(n) time where n is the length of the input list nums. The max function then iterates over the values of the counter, which in the worst case can also be O(n) if all numbers in nums are unique. However, since max() is working on the values and not the keys, and the values represent counts that can be at most n, the time for max() is O(u) where u is the number of unique numbers. Typically, u <= n, so the overall time complexity remains O(n).

The space complexity of the code is determined by the additional space required to store the elements counted by the Counter. In the worst case, if all elements in the array are unique, the Counter would need to store each unique element as a key with its associated count as a value, requiring O(u) space where u is the number of unique numbers in nums. Since u can be at most n, the space complexity is O(n).

Therefore, the reference answer is correct in stating that both the time complexity and space complexity of the code are O(n).

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