LeetCode 3069 - Distribute Elements Into Two Arrays I

LeetCode Problem 3069

Difficulty: 🟢 Easy
Topics: Array, Simulation

Solution

Problem Understanding

The problem is asking us to simulate a process where we distribute elements from a 1-indexed array of distinct integers nums into two separate arrays, arr1 and arr2, following a set of deterministic rules. Initially, the first element of nums goes to arr1, the second goes to arr2. Starting from the third element, the placement of nums[i] depends on comparing the last elements of the two arrays: if the last element of arr1 is greater than the last element of arr2, the next number goes to arr1; otherwise, it goes to arr2.

The output is the concatenation of arr1 followed by arr2, which forms the result array.

The constraints indicate that n is small (3 to 50), each number in nums is distinct and between 1 and 100. This guarantees no duplicates and a small enough input size for straightforward simulation. The important edge cases include sequences where numbers are strictly increasing, decreasing, or oscillating, because the comparison of last elements directly influences the distribution.

Approaches

The brute-force approach is a direct simulation. Start with empty arr1 and arr2. For each element in nums, append it according to the rules: the first goes to arr1, the second to arr2, and subsequent elements are appended based on the comparison of the last elements of arr1 and arr2. This is correct because it precisely follows the problem statement, and it is efficient enough given the constraints, since n <= 50.

No further optimization is needed because every element must be checked and placed in one of the arrays, so the process is inherently O(n). The key insight is recognizing that this is purely a simulation problem where following the rules step by step is sufficient.

Approach	Time Complexity	Space Complexity	Notes
Brute Force / Simulation	O(n)	O(n)	Directly simulates the rules described in the problem
Optimal	O(n)	O(n)	Same as brute force because simulation is already optimal

Algorithm Walkthrough

1. Initialize two empty arrays, arr1 and arr2.
2. Append the first element of nums to arr1 and the second element to arr2.
3. Iterate through the remaining elements of nums starting from index 2 (third element). For each element num:

• Compare the last elements of arr1 and arr2.
• If arr1[-1] > arr2[-1], append num to arr1.
• Otherwise, append num to arr2.

4. After the iteration, concatenate arr1 and arr2 to form result.
5. Return result.

Why it works: This works because the algorithm maintains the invariant that each new element is appended according to the last-element comparison rule. Since each element is placed exactly once, and concatenation preserves the order within arr1 and arr2, the resulting array accurately reflects the process described in the problem.

Python Solution

from typing import List

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

This implementation starts by initializing the first two elements in arr1 and arr2 as described. The loop iterates through the rest of nums, comparing the last elements and appending accordingly. Finally, the arrays are concatenated and returned.

Go Solution

func resultArray(nums []int) []int {
    arr1 := []int{nums[0]}
    arr2 := []int{nums[1]}
    
    for i := 2; i < len(nums); i++ {
        if arr1[len(arr1)-1] > arr2[len(arr2)-1] {
            arr1 = append(arr1, nums[i])
        } else {
            arr2 = append(arr2, nums[i])
        }
    }
    
    result := append(arr1, arr2...)
    return result
}

In Go, slices are used to store arr1 and arr2. The last element is accessed using len(arr1)-1 and len(arr2)-1. The append function handles slice resizing automatically. The arrays are concatenated with append(arr1, arr2...) before returning.

Worked Examples

Example 1: nums = [2,1,3]

Step	arr1	arr2	Operation
Initialize	[2]	[1]	first two elements
i=3	[2,3]	[1]	2 > 1, append 3 to arr1
Concatenate	[2,3,1]	-	result

Example 2: nums = [5,4,3,8]

Step	arr1	arr2	Operation
Initialize	[5]	[4]	first two elements
i=3	[5,3]	[4]	5 > 4, append 3 to arr1
i=4	[5,3]	[4,8]	3 < 4, append 8 to arr2
Concatenate	[5,3,4,8]	-	result

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each element is visited exactly once and placed in an array
Space	O(n)	Arrays arr1 and arr2 store all elements

The linear time and space are optimal because each element must be handled individually, and the arrays store all n elements.

Test Cases

# Provided examples
assert Solution().resultArray([2,1,3]) == [2,3,1]  # basic 3-element case
assert Solution().resultArray([5,4,3,8]) == [5,3,4,8]  # 4-element case

# Edge cases
assert Solution().resultArray([1,2,3,4,5]) == [1,3,5,2,4]  # strictly increasing
assert Solution().resultArray([5,4,3,2,1]) == [5,3,2,4,1]  # strictly decreasing
assert Solution().resultArray([10,1,20,2,30]) == [10,20,30,1,2]  # alternating high/low

# Minimum input size
assert Solution().resultArray([3,1,2]) == [3,2,1]  # n = 3

Test	Why
[2,1,3]	validates the simple 3-element scenario
[5,4,3,8]	tests mid-sized input and last-element comparison
[1,2,3,4,5]	strictly increasing sequence to check placement
[5,4,3,2,1]	strictly decreasing sequence to check placement
[10,1,20,2,30]	alternating highs and lows to stress comparison
[3,1,2]	minimum input size for correctness

Edge Cases

Strictly Increasing Array: If nums is strictly increasing, the smaller numbers keep going to arr2, while larger numbers may dominate arr1. This could cause an unbalanced split. The algorithm correctly compares last elements to maintain proper distribution.

Strictly Decreasing Array: Here, each new element tends to go to arr1 first, then arr2 if smaller than the last in arr2. The code handles this by always checking the last element comparison, ensuring correct placement.

Alternating High-Low Array: An input where elements alternate high and low values can test that the algorithm does not assume order beyond last-element comparison. The solution correctly appends to the proper array every time without assumptions.