We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsCompaniesDisclaimersPrivacyTermsContactAbout Us

Convert 1D Array Into 2D Array

Number: 2132

Difficulty: Easy

Paid? No

Companies: Google, Amazon, Meta

Problem Description

Given a 1D integer array named original and two integers m and n, construct a 2D array with m rows and n columns using all the elements from original. The first n elements form the first row, the next n elements form the second row, and so on. If the number of elements in original does not equal m * n, return an empty 2D array.

Key Insights

  • Ensure the total number of elements in original is exactly m * n; otherwise, constructing the desired 2D array is impossible.
  • Use slicing or index arithmetic to break original into chunks of n elements.
  • The simulation approach works by iteratively creating each row from the original array.

Space and Time Complexity

Time Complexity: O(m * n)
Space Complexity: O(m * n)

Solution

The solution checks whether the length of original matches m * n. If not, it immediately returns an empty 2D array. If the size requirement is met, iterate over the original array and slice it into segments of length n to form each row of the 2D array. This method leverages array slicing (or index arithmetic) to organize the elements correctly and build the required matrix.

Code Solutions

# Function to convert 1D array into 2D array
def construct2DArray(original, m, n):
    # Check if total elements equal m * n
    if len(original) != m * n:
        return []  # Return empty 2D array if condition is not met

    # Construct the 2D array by slicing original into chunks of n elements
    return [original[i * n:(i + 1) * n] for i in range(m)]

# Example usage:
print(construct2DArray([1, 2, 3, 4], 2, 2))  # Output: [[1, 2], [3, 4]]
← Back to All Questions