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Medium MatrixBFSDFS LeetCode

You get an n × n matrix of integers. Rotate the image 90 degrees clockwise. You must do it in place: do not allocate another n × n matrix for the final result (extra constant space for a few variables is fine).

Visual rule: what was the first row becomes the last column, and so on.

Example 1

  • Input: matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
  • Output: [[7, 4, 1], [8, 5, 2], [9, 6, 3]]

Example 2

  • Input: matrix = [[5, 1, 9, 11], [2, 4, 8, 10], [13, 3, 6, 7], [15, 14, 12, 16]]
  • Output: [[15, 13, 2, 5], [14, 3, 4, 1], [12, 6, 8, 9], [16, 7, 10, 11]]

Constraints

  • n == matrix.length == matrix[i].length
  • 1 <= n <= 20
  • -1000 <= matrix[i][j] <= 1000
Sample Input
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Sample Output
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Constraints
  • n == matrix.length == matrix[i].length
  • 1 <= n <= 20
  • -1000 <= matrix[i][j] <= 1000
Topics
MatrixBFSDFS

Transpose + Reverse Rows

function rotate(matrix) {
  const n = matrix.length;
  for (let i = 0; i < n; i++)
    for (let j = i + 1; j < n; j++)
      [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]];
  for (const row of matrix) row.reverse();
}

Time: O(n²) | Space: O(1)

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