How do you solve Distribute Elements Into Two Arrays I on LeetCode?

Distribute Elements Into Two Arrays I (LeetCode #3069) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: This approach leverages the same logic as the brute force but optimizes the way we handle the arrays by avoiding unnecessary checks and directly using the last elements of the arrays.

What algorithm or data structure is used to solve Distribute Elements Into Two Arrays I?

Distribute Elements Into Two Arrays I uses the following concepts: Array, Simulation. The recommended patterns to study are: Hash Map, Array.

Distribute Elements Into Two Arrays I — LeetCode #3069 (Easy)

Tags: Array, Simulation

Related patterns: Hash Map, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

In this approach, we simulate the process of appending elements to two arrays based on the conditions given. We simply iterate through the input array and decide where to append each element based on the last elements of the two arrays.

The time complexity is O(n²) because we are appending elements to the arrays and checking the last elements repeatedly, which can take linear time for each element in the worst case.

Step 1: Initialize two empty arrays, arr1 and arr2.
Step 2: Append the first element of nums to arr1 and the second element to arr2.
Step 3: For each subsequent element in nums, compare the last elements of arr1 and arr2 to decide where to append the current element.

1. Start with nums = [2, 1, 3]. arr1 = [], arr2 = []. 2. Append nums[0] (2) to arr1: arr1 = [2], arr2 = []. 3. Append nums[1] (1) to arr2: arr1 = [2], arr2 = [1]. 4. For i=2, compare arr1[-1] (2) and arr2[-1] (1). Since 2 > 1, append nums[2] (3) to arr1: arr1 = [2, 3], arr2 = [1]. 5. Concatenate: result = [2, 3, 1].

Optimal Solution approach

Time complexity: O(n). Space complexity: O(n).

This approach leverages the same logic as the brute force but optimizes the way we handle the arrays by avoiding unnecessary checks and directly using the last elements of the arrays.

The time complexity is O(n) because we only make a single pass through the input array, and the space complexity is O(n) due to the storage of the two result arrays.

Step 1: Initialize two empty arrays, arr1 and arr2.
Step 2: Append the first element of nums to arr1 and the second element to arr2.
Step 3: For each subsequent element in nums, directly compare the last elements of arr1 and arr2 to decide where to append the current element.

1. Start with nums = [5, 4, 3, 8]. arr1 = [], arr2 = []. 2. Append nums[0] (5) to arr1: arr1 = [5], arr2 = []. 3. Append nums[1] (4) to arr2: arr1 = [5], arr2 = [4]. 4. For i=2, compare arr1[-1] (5) and arr2[-1] (4). Since 5 > 4, append nums[2] (3) to arr1: arr1 = [5, 3], arr2 = [4]. 5. For i=3, compare arr1[-1] (3) and arr2[-1] (4). Since 4 > 3, append nums[3] (8) to arr2: arr1 = [5, 3], arr2 = [4, 8]. 6. Concatenate: result = [5, 3, 4, 8].

Common Mistakes

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#3069

Distribute Elements Into Two Arrays I

Easy

Array↗Simulation↗Hash Map↗Array↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

This approach leverages the same logic as the brute force but optimizes the way we handle the arrays by avoiding unnecessary checks and directly using the last elements of the arrays.

⚙️

Algorithm

3 steps

1Step 1: Initialize two empty arrays, arr1 and arr2.
2Step 2: Append the first element of nums to arr1 and the second element to arr2.
3Step 3: For each subsequent element in nums, directly compare the last elements of arr1 and arr2 to decide where to append the current element.

solution.py10 lines

1def distributeElements(nums):
2    arr1, arr2 = [], []
3    arr1.append(nums[0])
4    arr2.append(nums[1])
5    for i in range(2, len(nums)):
6        if arr1[-1] > arr2[-1]:
7            arr1.append(nums[i])
8        else:
9            arr2.append(nums[i])
10    return arr1 + arr2

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the input array, and the space complexity is O(n) due to the storage of the two result arrays.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.