Problem Description

You have a 1-indexed array nums containing distinct integers with length n. Your task is to split all elements of this array into two separate arrays (arr1 and arr2) following a specific set of rules.

The distribution process works as follows:

1. First operation: Put nums[1] into arr1
2. Second operation: Put nums[2] into arr2
3. For each subsequent operation (starting from the 3rd element):
   • Compare the last element of arr1 with the last element of arr2
   • If arr1[-1] > arr2[-1], append the current element to arr1
   • Otherwise, append the current element to arr2

After distributing all elements, you need to create a result array by concatenating arr1 and arr2 in that order.

Example: If after the distribution process arr1 = [1,2,3] and arr2 = [4,5,6], then result = [1,2,3,4,5,6].

The goal is to return this final result array after following the distribution rules for all elements in nums.

Intuition

The problem asks us to simulate a specific process of distributing elements between two arrays based on a comparison rule. Since we need to follow exact instructions for each operation, there's no room for optimization or clever tricks - we simply need to implement what's described.

The key insight is that this is a direct simulation problem. We're given explicit rules:

• Where to place the first two elements (fixed positions)
• A comparison-based rule for all subsequent elements

The decision for each element from index 3 onwards depends only on the current last elements of arr1 and arr2. This means we need to maintain these two arrays as we go, always having access to their last elements for comparison.

Since the problem requires us to return the concatenation of arr1 and arr2, and Python allows easy list concatenation with the + operator, we can directly return arr1 + arr2 after the distribution is complete.

The straightforward approach is to:

1. Initialize the two arrays with their mandatory first elements
2. Iterate through the remaining elements
3. Apply the comparison rule for each element
4. Concatenate and return

There's no need to overthink this problem - it's asking us to follow a recipe, so we implement exactly what the recipe says. The solution naturally flows from translating the problem statement into code.

Solution Approach

We implement a direct simulation following the problem's distribution rules.

Step 1: Initialize the two arrays

• Create arr1 with the first element: arr1 = [nums[0]]
• Create arr2 with the second element: arr2 = [nums[1]]

These initial placements are mandatory according to the problem rules.

Step 2: Process remaining elements Starting from index 2 (the third element in the 1-indexed array), we iterate through each element in nums:

for x in nums[2:]:

Step 3: Apply the comparison rule For each element x, we check the last elements of both arrays:

• If arr1[-1] > arr2[-1]: append x to arr1
• Otherwise: append x to arr2

if arr1[-1] > arr2[-1]:
    arr1.append(x)
else:
    arr2.append(x)

The -1 index gives us the last element of each array, which is what we need for the comparison.

Step 4: Concatenate and return After all elements have been distributed, we concatenate arr1 and arr2 using Python's list concatenation operator:

return arr1 + arr2

This creates a new list with all elements from arr1 followed by all elements from arr2.

The entire solution runs in O(n) time where n is the length of the input array, as we iterate through the array once and perform constant-time operations (comparison and append) for each element. The space complexity is also O(n) for storing the elements in arr1 and arr2.

Example Walkthrough

Let's walk through a concrete example with nums = [5, 4, 3, 6, 2, 7, 1].

Initial Setup:

• arr1 = [5] (first element goes to arr1)
• arr2 = [4] (second element goes to arr2)

Processing element 3 (index 2):

• Compare: arr1[-1] = 5 vs arr2[-1] = 4
• Since 5 > 4, add 3 to arr1
• arr1 = [5, 3], arr2 = [4]

Processing element 6 (index 3):

• Compare: arr1[-1] = 3 vs arr2[-1] = 4
• Since 3 ≤ 4, add 6 to arr2
• arr1 = [5, 3], arr2 = [4, 6]

Processing element 2 (index 4):

• Compare: arr1[-1] = 3 vs arr2[-1] = 6
• Since 3 ≤ 6, add 2 to arr2
• arr1 = [5, 3], arr2 = [4, 6, 2]

Processing element 7 (index 5):

• Compare: arr1[-1] = 3 vs arr2[-1] = 2
• Since 3 > 2, add 7 to arr1
• arr1 = [5, 3, 7], arr2 = [4, 6, 2]

Processing element 1 (index 6):

• Compare: arr1[-1] = 7 vs arr2[-1] = 2
• Since 7 > 2, add 1 to arr1
• arr1 = [5, 3, 7, 1], arr2 = [4, 6, 2]

Final Step:

• Concatenate: result = arr1 + arr2 = [5, 3, 7, 1, 4, 6, 2]

The key observation is that each element's placement depends solely on comparing the current last elements of the two arrays at that moment, creating a dynamic distribution pattern.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. The algorithm iterates through the array once starting from index 2, performing constant-time operations (comparison and append) for each element. Specifically, the loop runs n-2 times, and each iteration involves O(1) operations: comparing the last elements of arr1 and arr2, and appending to one of the arrays.

The space complexity is O(n), where n is the length of the array nums. The algorithm creates two new arrays arr1 and arr2 that together store all elements from the original array. Since every element from nums is distributed between these two arrays, the total space used is proportional to the input size. Additionally, the final concatenation arr1 + arr2 creates a new list of size n, but this doesn't change the overall space complexity which remains O(n).

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

Common Pitfalls

1. Index Out of Bounds Error with Small Arrays

A critical pitfall occurs when the input array has fewer than 2 elements. The code assumes nums has at least 2 elements when initializing arr1 = [nums[0]] and arr2 = [nums[1]]. If nums has only 0 or 1 element, this will raise an IndexError.

Solution: Add a guard clause to handle edge cases:

def resultArray(self, nums: List[int]) -> List[int]:
    # Handle edge cases
    if len(nums) == 0:
        return []
    if len(nums) == 1:
        return nums
  
    # Continue with normal logic
    arr1 = [nums[0]]
    arr2 = [nums[1]]
    # ... rest of the code

2. Misunderstanding the 1-Indexed Description

The problem states the array is "1-indexed" in the description, but the actual Python implementation uses 0-based indexing. This can cause confusion where developers might try to access nums[1] thinking it's the first element, when it's actually the second element in Python.

Solution: Be clear that while the problem description uses 1-based indexing for explanation, the implementation uses Python's standard 0-based indexing:

• nums[0] corresponds to the "first element" mentioned in the problem
• nums[1] corresponds to the "second element" mentioned in the problem

3. Inefficient Concatenation in Loops

If someone modifies the code to build the result array incrementally during the distribution process (rather than concatenating at the end), they might use repeated concatenation which creates new lists each time:

# Inefficient approach
result = []
for element in arr1:
    result = result + [element]  # Creates new list each time

Solution: Stick with the current approach of building separate arrays and concatenating once at the end, or use extend() method if building incrementally:

result = []
result.extend(arr1)
result.extend(arr2)

4. Assuming Elements Are Comparable

The code uses the > operator to compare elements. If the array contains mixed types that aren't comparable (though the problem specifies integers), this would raise a TypeError.

Solution: While not needed for this specific problem, defensive programming would validate input types if this were production code:

if not all(isinstance(x, (int, float)) for x in nums):
    raise ValueError("All elements must be numeric")