LeetCode #3069 — EASY

Distribute Elements Into Two Arrays I

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 1-indexed array of distinct integers nums of length n.

You need to distribute all the elements of nums between two arrays arr1 and arr2 using n operations. In the first operation, append nums[1] to arr1. In the second operation, append nums[2] to arr2. Afterwards, in the i^th operation:

If the last element of arr1 is greater than the last element of arr2, append nums[i] to arr1. Otherwise, append nums[i] to arr2.

The array result is formed by concatenating the arrays arr1 and arr2. For example, if arr1 == [1,2,3] and arr2 == [4,5,6], then result = [1,2,3,4,5,6].

Return the array result.

Example 1:

Input: nums = [2,1,3]
Output: [2,3,1]
Explanation: After the first 2 operations, arr1 = [2] and arr2 = [1].
In the 3^rd 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]
Explanation: After the first 2 operations, arr1 = [5] and arr2 = [4].
In the 3^rd 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 4^th 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.

Step 01

Brute Force Baseline

Problem summary: You are given a 1-indexed array of distinct integers nums of length n. You need to distribute all the elements of nums between two arrays arr1 and arr2 using n operations. In the first operation, append nums[1] to arr1. In the second operation, append nums[2] to arr2. Afterwards, in the ith operation: If the last element of arr1 is greater than the last element of arr2, append nums[i] to arr1. Otherwise, append nums[i] to arr2. The array result is formed by concatenating the arrays arr1 and arr2. For example, if arr1 == [1,2,3] and arr2 == [4,5,6], then result = [1,2,3,4,5,6]. Return the array result.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[2,1,3]

Example 2

[5,4,3,8]

Core Insight

What unlocks the optimal approach

Divide the array into two arrays by keeping track of the last elements of both subarrays.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3069: Distribute Elements Into Two Arrays I
class Solution {
    public int[] resultArray(int[] nums) {
        int n = nums.length;
        int[] arr1 = new int[n];
        int[] arr2 = new int[n];
        arr1[0] = nums[0];
        arr2[0] = nums[1];
        int i = 0, j = 0;
        for (int k = 2; k < n; ++k) {
            if (arr1[i] > arr2[j]) {
                arr1[++i] = nums[k];
            } else {
                arr2[++j] = nums[k];
            }
        }
        for (int k = 0; k <= j; ++k) {
            arr1[++i] = arr2[k];
        }
        return arr1;
    }
}

// Accepted solution for LeetCode #3069: Distribute Elements Into Two Arrays I
func resultArray(nums []int) []int {
	arr1 := []int{nums[0]}
	arr2 := []int{nums[1]}
	for _, x := range nums[2:] {
		if arr1[len(arr1)-1] > arr2[len(arr2)-1] {
			arr1 = append(arr1, x)
		} else {
			arr2 = append(arr2, x)
		}
	}
	return append(arr1, arr2...)
}

# Accepted solution for LeetCode #3069: Distribute Elements Into Two Arrays I
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

// Accepted solution for LeetCode #3069: Distribute Elements Into Two Arrays I
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3069: Distribute Elements Into Two Arrays I
// class Solution {
//     public int[] resultArray(int[] nums) {
//         int n = nums.length;
//         int[] arr1 = new int[n];
//         int[] arr2 = new int[n];
//         arr1[0] = nums[0];
//         arr2[0] = nums[1];
//         int i = 0, j = 0;
//         for (int k = 2; k < n; ++k) {
//             if (arr1[i] > arr2[j]) {
//                 arr1[++i] = nums[k];
//             } else {
//                 arr2[++j] = nums[k];
//             }
//         }
//         for (int k = 0; k <= j; ++k) {
//             arr1[++i] = arr2[k];
//         }
//         return arr1;
//     }
// }

// Accepted solution for LeetCode #3069: Distribute Elements Into Two Arrays I
function resultArray(nums: number[]): number[] {
    const arr1: number[] = [nums[0]];
    const arr2: number[] = [nums[1]];
    for (const x of nums.slice(2)) {
        if (arr1.at(-1)! > arr2.at(-1)!) {
            arr1.push(x);
        } else {
            arr2.push(x);
        }
    }
    return arr1.concat(arr2);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.