Matrix Rotate - FreeCodeCamp #27 Daily Challenge

Matrix Rotate - FreeCodeCamp #27

🚩 Problem Statement

Given a matrix (array of arrays), rotate it 90° clockwise. Example: for [[1,2],[3,4]], the output should be [[3,1],[4,2]].

1. Initial Analysis

What does it mean to rotate a matrix 90°? Each row of the original becomes a column in the new one, but in reverse order. Visualize it like this:

flowchart TD
  A["[1,2]"] --> C["[3,1]"]
  B["[3,4]"] --> D["[4,2]"]

Key Test Cases

• 1x1: Input: [[1]] → Output: [[1]]
• 2x2: Input: [[1,2],[3,4]] → Output: [[3,1],[4,2]]
• 3x3: Input: [[1,2,3],[4,5,6],[7,8,9]] → Output: [[7,4,1],[8,5,2],[9,6,3]]
• With zeros: Input: [[0,1,0],[1,0,1],[0,0,0]] → Output: [[0,1,0],[0,0,1],[0,1,0]}

2. Strategy & Step by Step

How do we solve it? The clearest way is to create a new empty matrix and map each element to its new position. The trick is in the indices:

• The element at (i, j) in the original goes to (j, n-1-i) in the rotated, where n is the number of rows.

Algorithm:

1. Get dimensions: rows n, columns m.
2. Create an empty matrix of size m x n.
3. For each element (i, j), place it at (j, n-1-i).
4. Return the rotated matrix.

3. JavaScript Implementation

function rotate(matrix) {
  if (!matrix.length || !matrix[0].length)
    return []
  const n = matrix.length
  const m = matrix[0].length
  const rotated = Array.from({ length: m }, () => new Array(n))
  for (let i = 0; i < n; i++) {
    for (let j = 0; j < m; j++) {
      rotated[j][n - 1 - i] = matrix[i][j]
    }
  }
  return rotated
}

4. Complexity & Edge Cases

Complexity

• Time: $O(n \times m)$ (we visit every element)
• Space: $O(n \times m)$ (new matrix)

Edge Cases

• Empty matrix: [] → []
• Empty rows: [[ ]] → []
• 1x1: remains the same
• Non-square: works for any shape

5. Reflections & Learnings

What did we learn?

• Index manipulation and 2D arrays
• Nested loops and position mapping
• Input validation and edge cases
• Complexity analysis

Can it be optimized?

If the matrix is square and you can modify it, you can rotate in-place: first transpose, then reverse each row. But for rectangular matrices or if you want to preserve the original, the presented method is the safest and clearest.

6. Resources

• MDN Arrays
• GeeksForGeeks: Rotate matrix
• LeetCode: Rotate Image

Ready to try with a bigger matrix? What if the matrix is very rectangular? 🤔