How do you solve Transpose Matrix on LeetCode?

Transpose Matrix (LeetCode #867) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: Understanding the concept of transposing a matrix is crucial.

What is the brute force solution for Transpose Matrix?

The brute force approach for Transpose Matrix is Brute Force, with O(m * n) time complexity and O(m * n) space complexity. The brute-force approach involves creating a new matrix where we manually assign values from the original matrix by swapping rows and columns. This is straightforward and helps us understand the concept of transposing a matrix. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Transpose Matrix?

Transpose Matrix uses the following concepts: Array, Matrix, Simulation. The recommended patterns to study are: Array, Matrix.

What are the interview tips for Transpose Matrix?

Always clarify the dimensions of the input matrix. Discuss the approach before coding to ensure understanding. Consider edge cases like single-row or single-column matrices.

What are common mistakes when solving Transpose Matrix?

Not initializing the new matrix with the correct dimensions. Confusing row and column indices during assignment.

#867

Transpose Matrix

Easy

ArrayMatrixSimulationArrayMatrix

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n)	O(m * n)
Space	O(m * n)	O(m * n)

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Intuition

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

The optimal approach leverages the properties of the transpose operation and can be implemented efficiently using a single loop with in-place swapping when modifying the original matrix. However, since we need a new matrix, we will still create a new one but with a clear understanding of the dimensions.

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Algorithm

2 steps

1Step 1: Create a new matrix with dimensions swapped (n x m).
2Step 2: Use a single loop to fill in the new matrix by accessing elements from the original matrix.

solution.py7 lines

1def transpose(matrix):
2    m, n = len(matrix), len(matrix[0])
3    transposed = [[0] * m for _ in range(n)]
4    for i in range(m):
5        for j in range(n):
6            transposed[j][i] = matrix[i][j]
7    return transposed

ℹ

Complexity note: The time complexity remains O(m * n) as we still need to visit each element. The space complexity is also O(m * n) due to the new matrix.

1Understanding the concept of transposing a matrix is crucial.
2Recognizing that the new matrix dimensions are swapped helps in implementation.

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