#3069

Easy

LeetCode Problem Workspace

Distribute Elements Into Two Arrays I

Distribute elements from a distinct integer array into two subarrays using a sequential simulation strategy for optimal ordering.

array simulation

Problem Statement

You are given a 1-indexed array of distinct integers nums of length n. Your task is to distribute the elements into two separate arrays, arr1 and arr2, over n sequential operations, starting by placing the first element into arr1 and the second into arr2. For each following element, you must decide which subarray to append it to based on the relative size of the last elements in arr1 and arr2.

After distributing all elements, form the final result array by concatenating arr1 followed by arr2. For example, if arr1 = [1,2,3] and arr2 = [4,5,6], the resulting array will be [1,2,3,4,5,6]. Your goal is to perform this array simulation correctly to produce the expected output for any given nums array.

Examples

Example 1

Input: nums = [2,1,3]

Output: [2,3,1]

After the first 2 operations, arr1 = [2] and arr2 = [1]. In the 3rd operation, as the last element of arr1 is greater than the last element of arr2 (2 > 1), append nums[3] to arr1. After 3 operations, arr1 = [2,3] and arr2 = [1]. Hence, the array result formed by concatenation is [2,3,1].

Example 2

Input: nums = [5,4,3,8]

Output: [5,3,4,8]

After the first 2 operations, arr1 = [5] and arr2 = [4]. In the 3rd operation, as the last element of arr1 is greater than the last element of arr2 (5 > 4), append nums[3] to arr1, hence arr1 becomes [5,3]. In the 4th operation, as the last element of arr2 is greater than the last element of arr1 (4 > 3), append nums[4] to arr2, hence arr2 becomes [4,8]. After 4 operations, arr1 = [5,3] and arr2 = [4,8]. Hence, the array result formed by concatenation is [5,3,4,8].

Constraints

3 <= n <= 50
1 <= nums[i] <= 100
All elements in nums are distinct.

Solution Approach

Initialize Two Subarrays

Begin by assigning nums[0] to arr1 and nums[1] to arr2. This establishes the simulation starting points and provides reference for subsequent comparisons.

Sequentially Append Elements

For each element from the third to the last, compare it with the last elements of arr1 and arr2. Append to arr1 if its last element is greater, otherwise append to arr2. This ensures the simulation respects the problem's distribution rules.

Concatenate to Form Result

Once all elements are placed, concatenate arr1 and arr2 to create the final result array. This step completes the simulation and produces the expected linear output.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n) because each element is processed once, and space complexity is O(n) to store the two subarrays before concatenation.

What Interviewers Usually Probe

Ask why comparing only the last elements of arr1 and arr2 is sufficient.
Check if the candidate correctly simulates sequential placement rather than sorting.
Probe for understanding of array concatenation versus in-place merging.

Common Pitfalls or Variants

Common pitfalls

Forgetting that arrays are 1-indexed and misplacing initial elements.
Incorrectly comparing the current element with both subarrays instead of only the last elements.
Concatenating arr2 before arr1, which reverses the required order.

Follow-up variants

Distribute elements using maximum element comparison instead of last-element simulation.
Extend the problem to allow repeated elements and handle equal comparisons.
Simulate distribution for k subarrays instead of only two, keeping track of last elements.

FAQ

What is the main pattern behind Distribute Elements Into Two Arrays I?

The pattern is Array plus Simulation, where each element is appended based on a sequential comparison with the last elements of two subarrays.

Can this problem handle arrays with repeated elements?

No, the problem constraints require all elements in nums to be distinct for correct simulation behavior.

Why is comparing only the last elements enough?

Because the sequential placement ensures the ordering within each subarray is maintained, making further comparisons unnecessary.

What is the maximum input size allowed?

The array length n can range from 3 to 50 as per constraints.

Is it possible to solve this without creating two subarrays?

Yes, but using two subarrays makes the simulation clearer and avoids off-by-one placement errors in concatenation.

terminal

Solution

Solution 1: Simulation

We create two arrays $\textit{arr1}$ and $\textit{arr2}$ to store the elements of $\textit{nums}$. Initially, $\textit{arr1}$ contains only $\textit{nums[0]}$, and $\textit{arr2}$ contains only $\textit{nums[1]}$.

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class Solution:
    def resultArray(self, nums: List[int]) -> List[int]:
        arr1 = [nums[0]]
        arr2 = [nums[1]]
        for x in nums[2:]:
            if arr1[-1] > arr2[-1]:
                arr1.append(x)
            else:
                arr2.append(x)
        return arr1 + arr2

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