Diagonal Traverse

mediumFREEArrays & Matrix OperationsDSA for AI Engineers

Given an m x n matrix, return all elements in diagonal order (alternating up-right and down-left), starting at the top-left element.

Return the elements as a list. The test harness prints them space-separated.

Example

Input

1,2,3
4,5,6
7,8,9

Output

1 2 4 7 5 3 6 8 9

Reasoning

The input is a $3 \times 3$ grid read row by row: 1,2,3 is row 0, 4,5,6 is row 1, 7,8,9 is row 2.
An anti-diagonal is a set of cells running from upper-right to lower-left. A $3 \times 3$ grid has $3 + 3 - 1 = 5$ of them, holding $\{1\}$ , $\{2, 4\}$ , $\{3, 5, 7\}$ , $\{6, 8\}$ and $\{9\}$ .
The first one is just $1$ , the top-left corner. The second is read moving down-left, so $2$ comes before $4$ .
The third is read moving up-right: it starts at the bottom-left $7$ , passes the centre $5$ and ends at the top-right $3$ .
The fourth flips back to down-left ( $6$ then $8$ ), and the last is the corner $9$ .
Joining them gives 1 2 4 7 5 3 6 8 9. Notice that every element appears exactly once, and that the reading direction reverses on each successive diagonal. For a $2 \times 2$ grid with rows $1, 2$ and $3, 4$ the same pattern happens to give 1 2 3 4.

Constraints

1 <= m, n <= 10^4
1 <= m * n <= 10^4

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Concept primer

Background Knowledge

The problem of "Diagonal Traverse" involves traversing a 2D matrix in a diagonal order, which alternates between up-right and down-left directions. To understand this problem, it's essential to have a solid grasp of matrix operations and array indexing. In a 2D matrix, each element is identified by its row and column indices. The diagonal order traversal requires navigating through the matrix in a way that visits elements in a diagonal pattern.

Key concepts to understand include matrix representation, where a matrix is represented as a collection of rows and columns, and indexing, which refers to accessing specific elements within the matrix using their row and column indices. Additionally, understanding the difference between row-major and column-major ordering is crucial, as it affects how elements are stored and accessed in the matrix.

In the context of diagonal order traversal, it's essential to recognize that the diagonal pattern can start from the top-left corner and move down-right or start from the top-right corner and move down-left. The problem requires alternating between these two diagonal directions, which involves careful management of row and column indices to ensure correct traversal.

Algorithm/Approach

The algorithmic approach to solving this problem involves using a combination of looping constructs and conditional statements to navigate the matrix in a diagonal order. The general pattern involves iterating over the matrix elements, using the row and column indices to determine the current diagonal direction, and adjusting the indices accordingly to move to the next element in the diagonal order.

Step-by-Step Strategy

To implement the solution, follow these steps:

Initialize variables to keep track of the current row and column indices.
Determine the starting point of the diagonal traversal (e.g., top-left or top-right corner).
Use a looping construct to iterate over the matrix elements, adjusting the row and column indices based on the current diagonal direction.
Use conditional statements to alternate between up-right and down-left diagonal directions.
Store the visited elements in a data structure (e.g., array or list) to output the diagonal order.

Common Pitfalls

When implementing the solution, watch out for:

Incorrect indexing or bounds checking, which can lead to index out of range errors.
Failure to alternate between diagonal directions, resulting in an incorrect traversal order.
Inefficient use of data structures or looping constructs, which can impact performance.

Time & Space Complexity

The expected time complexity for this problem is $O(m \cdot n)$ , where $m$ and $n$ are the dimensions of the input matrix, since we need to visit each element once. The space complexity is also $O(m \cdot n)$ , as we need to store the visited elements in a data structure. However, the actual space complexity may vary depending on the specific implementation and data structures used.

solution.py

Python 3.13

Test results