Diagonal Traverse

Diagonal Traverse

What is this problem about?

The Diagonal Traverse interview question asks you to return all elements of an $m imes n$ matrix in diagonal order. You start at (0,0) and move diagonally upwards, then switch to diagonally downwards for the next diagonal, and so on. The movement zig-zags through the matrix until every cell is visited.

Why is this asked in interviews?

Companies like Google, Microsoft, and Bloomberg frequently use this problem to test a candidate's ability to handle complex 2D array traversal logic. It evaluations how you manage boundary conditions and switch states (directions). It’s a test of simulation interview patterns and your ability to maintain multiple pointers or state variables simultaneously.

Algorithmic pattern used

The most common pattern is Simulation with a direction flag.

1. Notice that for any cell $(r, c)$, the sum $r + c$ is constant for all cells on the same diagonal.
2. Iterate through all possible sums from $0$ to $m + n - 2$.
3. For each sum, collect all cells where $r + c == sum$.
4. If the sum is even, traverse upwards; if odd, traverse downwards (or vice-versa, depending on the problem's specific zig-zag start).

Example explanation

Matrix $3 imes 3$:

1 2 3
4 5 6
7 8 9

1. Sum 0: (0,0) $o$ [1].
2. Sum 1: (0,1), (1,0). Upward? No, next is downward. $o$ [2, 4].
3. Sum 2: (2,0), (1,1), (0,2). Upward $o$ [7, 5, 3]. Result: [1, 2, 4, 7, 5, 3, 6, 8, 9]. (Wait, actual zig-zag usually starts up: 1, 2, 4... no, typically 1, 2, 4, 7, 5, 3, 6, 8, 9 is one variation, another is 1, 2, 4, 7, 5, 3... just follow the sum logic).

Common mistakes candidates make

• Boundary Overflow: Forgetting to check if $r < m$ and $c < n$ when calculating the next coordinate.
• Direction Flip: Switching the direction at the wrong time or failing to identify the correct starting point for each diagonal.
• Inefficient collection: Using a nested loop that visits every cell multiple times instead of a single pass approach.

Interview preparation tip

The key invariant in diagonal problems is $r + c = constant$. For anti-diagonals, it's $r - c = constant$. Remembering these two properties makes any matrix diagonal problem significantly easier.