Problem Description

You are given a positive integer array nums. Your task is to split this array into two non-empty arrays, nums1 and nums2, such that every element from the original array belongs to exactly one of these two arrays.

The partition value is calculated as |max(nums1) - min(nums2)|, where:

• max(nums1) is the largest element in nums1
• min(nums2) is the smallest element in nums2
• |...| denotes the absolute value

Your goal is to find a way to partition the array such that this partition value is minimized, and return this minimum partition value.

For example, if you have an array [1, 2, 3, 4], you could partition it as nums1 = [1, 2] and nums2 = [3, 4]. The partition value would be |max([1, 2]) - min([3, 4])| = |2 - 3| = 1.

The key insight is that to minimize |max(nums1) - min(nums2)|, we want the maximum of nums1 and the minimum of nums2 to be as close as possible. After sorting the array, the optimal partition can be found by checking adjacent elements - one serving as the maximum of nums1 and the other as the minimum of nums2. The solution iterates through all adjacent pairs in the sorted array and returns the minimum difference found.

Intuition

Let's think about what we're trying to minimize: |max(nums1) - min(nums2)|. This value represents the gap between the largest element in the first partition and the smallest element in the second partition.

To minimize this gap, we want these two values to be as close as possible. Here's the key observation: imagine we have sorted the array. If we pick any element as max(nums1), then all elements smaller than or equal to it could potentially be in nums1, while all elements larger than it must be in nums2.

Since we want to minimize the difference between max(nums1) and min(nums2), and min(nums2) must be larger than max(nums1) (otherwise that element could be in nums1 instead), the best choice for min(nums2) would be the smallest element that's larger than max(nums1).

In a sorted array, the smallest element larger than any given element is simply the next element in the sequence. This means that the optimal partition will always have max(nums1) and min(nums2) as consecutive elements in the sorted array.

For example, if we have sorted array [1, 3, 5, 9], we could:

• Set max(nums1) = 1 and min(nums2) = 3, giving us difference 3 - 1 = 2
• Set max(nums1) = 3 and min(nums2) = 5, giving us difference 5 - 3 = 2
• Set max(nums1) = 5 and min(nums2) = 9, giving us difference 9 - 5 = 4

The minimum difference is 2, which occurs between consecutive elements. This is why the solution simply sorts the array and finds the minimum difference between adjacent elements - this represents the optimal way to partition the array.

Learn more about Sorting patterns.

Solution Approach

The implementation follows a straightforward approach based on our intuition that the optimal partition occurs at consecutive elements in a sorted array.

Step 1: Sort the array

nums.sort()

We first sort the array in ascending order. This allows us to easily identify which elements should be adjacent to minimize the partition value. Sorting takes O(n log n) time complexity.

Step 2: Find minimum difference between adjacent elements

return min(b - a for a, b in pairwise(nums))

After sorting, we iterate through all consecutive pairs of elements using the pairwise() function. The pairwise() function creates pairs of consecutive elements: (nums[0], nums[1]), (nums[1], nums[2]), ..., (nums[n-2], nums[n-1]).

For each pair (a, b) where a < b:

• a represents the maximum element of nums1
• b represents the minimum element of nums2
• The partition value for this split is b - a (no absolute value needed since b > a in sorted array)

We calculate the difference b - a for each consecutive pair and return the minimum among all these differences.

Example walkthrough: For nums = [5, 1, 9, 3]:

1. After sorting: [1, 3, 5, 9]
2. Consecutive pairs and their differences:
   • (1, 3): difference = 3 - 1 = 2
   • (3, 5): difference = 5 - 3 = 2
   • (5, 9): difference = 9 - 5 = 4
3. Return minimum difference = 2

The time complexity is O(n log n) due to sorting, and the space complexity is O(1) if we don't count the space used by sorting algorithm.

Example Walkthrough

Let's walk through the solution with nums = [7, 2, 10, 4, 1]:

Step 1: Sort the array

• Original: [7, 2, 10, 4, 1]
• Sorted: [1, 2, 4, 7, 10]

Step 2: Check all possible partitions using adjacent elements

For each adjacent pair, the left element becomes max(nums1) and the right element becomes min(nums2):

1. Partition at (1, 2):

   • nums1 = [1], nums2 = [2, 4, 7, 10]
   • max(nums1) = 1, min(nums2) = 2
   • Partition value = |1 - 2| = 1

2. Partition at (2, 4):

   • nums1 = [1, 2], nums2 = [4, 7, 10]
   • max(nums1) = 2, min(nums2) = 4
   • Partition value = |2 - 4| = 2

3. Partition at (4, 7):

   • nums1 = [1, 2, 4], nums2 = [7, 10]
   • max(nums1) = 4, min(nums2) = 7
   • Partition value = |4 - 7| = 3

4. Partition at (7, 10):

   • nums1 = [1, 2, 4, 7], nums2 = [10]
   • max(nums1) = 7, min(nums2) = 10
   • Partition value = |7 - 10| = 3

Step 3: Return the minimum partition value

• All partition values: [1, 2, 3, 3]
• Minimum = 1

The optimal partition is nums1 = [1] and nums2 = [2, 4, 7, 10] with a partition value of 1.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log n)

The time complexity is dominated by the sorting operation nums.sort(), which takes O(n × log n) time where n is the length of the input array. After sorting, the code iterates through adjacent pairs using pairwise(nums) to find the minimum difference, which takes O(n) time. Since O(n × log n) + O(n) = O(n × log n), the overall time complexity is O(n × log n).

Space Complexity: O(log n)

The space complexity comes from the sorting algorithm. Python's built-in sort() method uses Timsort, which requires O(log n) space for the recursive call stack in the worst case. The pairwise iterator and the variable storing the minimum value use only O(1) additional space. Therefore, the total space complexity is O(log n).

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

Common Pitfalls

1. Forgetting to sort the array

One of the most common mistakes is attempting to find the minimum partition value without first sorting the array. Without sorting, checking adjacent elements in the original array won't guarantee the optimal partition.

Incorrect approach:

def findValueOfPartition(self, nums: List[int]) -> int:
    min_difference = float('inf')
    for i in range(1, len(nums)):
        # This won't work with unsorted array!
        current_difference = abs(nums[i] - nums[i - 1])
        min_difference = min(min_difference, current_difference)
    return min_difference

Why it fails: For nums = [5, 1, 9, 3], this would check pairs (5,1), (1,9), (9,3) instead of the sorted pairs (1,3), (3,5), (5,9), potentially missing the optimal partition.

2. Misunderstanding the partition requirement

Some might think we need to maintain the original order or create contiguous subarrays, leading to overly complex solutions.

Incorrect interpretation:

def findValueOfPartition(self, nums: List[int]) -> int:
    min_val = float('inf')
    # Trying to split at each position maintaining order
    for i in range(1, len(nums)):
        left = nums[:i]
        right = nums[i:]
        min_val = min(min_val, abs(max(left) - min(right)))
    return min_val

Solution: Remember that elements can be placed in any order within nums1 and nums2. The only requirement is that every element belongs to exactly one partition.

3. Using absolute value unnecessarily after sorting

After sorting, we know that nums[i] < nums[i+1], so nums[i+1] - nums[i] is always positive. Using abs() adds unnecessary computation.

Redundant code:

min_difference = min(min_difference, abs(nums[i] - nums[i - 1]))  # abs() not needed

Better approach:

min_difference = min(min_difference, nums[i] - nums[i - 1])

4. Incorrect boundary handling

Starting the loop from index 0 instead of 1, or going up to len(nums) instead of len(nums)-1 when using nums[i+1].

Off-by-one error:

# Wrong: starts from 0
for i in range(len(nums) - 1):
    current_difference = nums[i] - nums[i - 1]  # nums[-1] when i=0!

Correct approach:

# Start from 1 when using nums[i-1]
for i in range(1, len(nums)):
    current_difference = nums[i] - nums[i - 1]

5. Not handling edge cases

While the problem states we need two non-empty arrays, forgetting to consider minimum array size (n ≥ 2) could lead to errors.

Defensive programming:

def findValueOfPartition(self, nums: List[int]) -> int:
    if len(nums) < 2:
        return 0  # or raise an exception
  
    nums.sort()
    min_difference = float('inf')
    for i in range(1, len(nums)):
        min_difference = min(min_difference, nums[i] - nums[i - 1])
    return min_difference