How do you solve Rotate Image on LeetCode?

Rotate Image (LeetCode #48) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(1) space complexity. Key insight: Rotating a matrix can be achieved through transposition and row reversal.

What is the brute force solution for Rotate Image?

The brute force approach for Rotate Image is Brute Force, with O(n²) time complexity and O(n²) space complexity. The brute-force approach involves creating a new matrix to hold the rotated values. We calculate the new position for each element and fill the new matrix accordingly. Finally, we copy the new matrix back to the original. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Rotate Image?

Rotate Image uses the following concepts: Array, Math, Matrix. The recommended patterns to study are: Matrix Manipulation, In-Place Algorithms.

What are the interview tips for Rotate Image?

Clearly explain your thought process before coding. Consider edge cases, such as the smallest matrix (1x1). Practice both brute-force and optimal solutions to understand trade-offs.

What are common mistakes when solving Rotate Image?

Not understanding the need for in-place rotation and using extra space. Confusing the transposition step with the final rotation.

#48

Rotate Image

Medium

ArrayMathMatrixMatrix ManipulationIn-Place Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n²)Space O(1)

The optimal solution uses a two-step process: first, we transpose the matrix, and then we reverse each row. This allows us to rotate the matrix in-place without using extra space.

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Algorithm

2 steps

1Step 1: Transpose the matrix by swapping matrix[i][j] with matrix[j][i].
2Step 2: Reverse each row of the transposed matrix.

solution.py9 lines

1def rotate(matrix):
2    n = len(matrix)
3    # Transpose
4    for i in range(n):
5        for j in range(i + 1, n):
6            matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
7    # Reverse each row
8    for i in range(n):
9        matrix[i].reverse()

ℹ

Complexity note: The time complexity remains O(n²) due to the two nested loops for transposing and reversing. The space complexity is O(1) because we are modifying the matrix in place.

1Rotating a matrix can be achieved through transposition and row reversal.
2In-place operations are crucial for optimizing space complexity.

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