The square case is pretty mechanical once you realize you only need to swap the upper triangle with the lower triangle.
First, explain the in-place transpose algorithm for a square matrix by swapping elements across the diagonal, emphasizing the O(n^2) time and O(1) space. Then, discuss how to handle a non-square matrix, either by creating a new matrix or by using a different in-place approach if possible, and analyze the complexity changes.
Pro tip: Mention that for non-square matrices, in-place transpose is not possible without additional space due to the different dimensions, and that the time complexity remains O(m*n) but space becomes O(m*n) for a new matrix. This shows you understand the trade-offs.
Restate the problem: transpose an n x n matrix in place with O(n^2) time and O(1) space. Confirm that 'in place' means modifying the original matrix without using extra space proportional to n.
Describe iterating over the upper triangle (i from 0 to n-1, j from i+1 to n-1) and swapping matrix[i][j] with matrix[j][i]. This achieves transpose in O(n^2) time and O(1) space.
State that the number of swaps is n(n-1)/2, which is O(n^2), and no extra space is used, so O(1) space.
For an m x n matrix, the transpose is n x m. In-place is not possible because the dimensions change. You must allocate a new n x m matrix and copy elements: new[j][i] = old[i][j]. Time remains O(m*n), but space becomes O(m*n).
Highlight that time complexity stays O(m*n) but space increases to O(m*n). If in-place is required, consider if the matrix can be represented as a 1D array and transposed using index arithmetic, but that still requires O(m*n) space for the new array unless the matrix is square.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.