How do you solve Convert 1D Array Into 2D Array on LeetCode?

Convert 1D Array Into 2D Array (LeetCode #2022) 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: The total number of elements must match the dimensions of the desired 2D array.

What is the brute force solution for Convert 1D Array Into 2D Array?

The brute force approach for Convert 1D Array Into 2D Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking if we can fill the 2D array by iterating through the original array and manually placing elements into the new structure. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Convert 1D Array Into 2D Array?

Convert 1D Array Into 2D Array uses the following concepts: Array, Matrix, Simulation. The recommended patterns to study are: Array, Matrix.

What are the interview tips for Convert 1D Array Into 2D Array?

Always validate input conditions before processing. Think about how to optimize your solution by reducing unnecessary iterations. Practice explaining your thought process clearly as you code.

What are common mistakes when solving Convert 1D Array Into 2D Array?

Forgetting to check if m * n equals the length of the original array before proceeding. Using nested loops unnecessarily, which can lead to inefficient solutions.

#2022

Convert 1D Array Into 2D Array

Easy

ArrayMatrixSimulationArrayMatrix

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)

The optimal approach leverages the fact that we can directly compute the indices for the 2D array without needing nested loops. This reduces unnecessary complexity and improves performance.

⚙️

Algorithm

3 steps

1Step 1: Check if m * n is equal to the length of the original array. If not, return an empty array.
2Step 2: Create a new 2D array of size m x n.
3Step 3: Use a single loop to fill the 2D array by calculating the row and column indices directly.

solution.py9 lines

1# Full working Python code
2
3def construct2DArray(original, m, n):
4    if m * n != len(original):
5        return []
6    result = [[0] * n for _ in range(m)]
7    for i in range(len(original)):
8        result[i // n][i % n] = original[i]
9    return result

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the original array once. The space complexity is also O(n) due to the new 2D array created.

1The total number of elements must match the dimensions of the desired 2D array.
2Directly calculating indices can simplify the logic and improve performance.

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