The O(n) Club: Spiral Matrix II: Mastering The Swirly-Square Algorithm Without Losing Your Sanity

The O(n) Club: Spiral Matrix II — Mastering The Swirly-Square Algorithm Without Losing Your Sanity

⚡ TL;DR

Fill an n x n grid in spiral order with numbers 1 to n²: right, down, left, up, repeat until you’ve filled every cell—or collapsed from existential dread. Brute-force means juggling boundaries like flaming torches. Here’s a Java taste:

// Quick spiral fill
int[][] matrix = new int[n][n];
int left = 0, right = n - 1, top = 0, bottom = n - 1, num = 1;
while (num <= n * n) {
    for (int i = left; i <= right; i++) matrix[top][i] = num++;
    top++;
    for (int i = top; i <= bottom; i++) matrix[i][right] = num++;
    right--;
    for (int i = right; i >= left; i--) matrix[bottom][i] = num++;
    bottom--;
    for (int i = bottom; i >= top; i--) matrix[i][left] = num++;
    left++;
}

🧠 How Devs Usually Mess This Up

Let’s be real: most devs launch into Spiral Matrix II expecting a pleasant drive, but instead end up skidding off the edge at Boundary Off-By-One Pass. Classic rookie goofs include:

• Obsessively writing four almost-identical for-loops per ring (welcome to loop purgatory—no one gets out alive).
• Off-by-one brainfarts that double-fill the poor, innocent center cell or, worse, miss it completely.
• Mixing up Spiral Matrix I and II, like confusing reading IKEA instructions (I) versus performing actual Swedish hex-wrench jazz (II).
• Forgetting to adjust every boundary, causing your matrix to spiral into the nether realms of Floyd-Warshall weirdness.

🔍 What’s Actually Going On

Pretend you’re a robot chef decorating a cake in the fanciest spiral possible: start at the upper left, pipe right, curve down the side, swing left, scamper up, and just keep zipping inward until you run out of buttercream or, in our case, cells. The only real trick? Shrink your bounds each lap (top, right, bottom, left) so you don’t decorate the same spot twice.

Or, if you prefer your code with some vector spice, use direction deltas like {0,1}, {1,0}, {0,-1}, {-1,0} and change direction whenever you’re about to wander off-grid or bump into a pre-filled square (no, Java won’t power up Pac-Man mode for you—manual labor only).

🛠️ PseudoCode

• Initialize the Matrix:

  int[][] matrix = new int[n][n];

  Start with zeros—Java’s default, your brain’s relief.
• Set Boundaries:

  int top = 0, bottom = n - 1, left = 0, right = n - 1;

  These boundaries keep your spiral from going full Da Vinci.
• Fill in Spiral Order:

  int num = 1;
  while (num <= n * n) {
     // Fill top row left → right
     // Fill right col top → bottom
     // Fill bottom row right → left
     // Fill left col bottom → top
     // Update boundaries

  Update each wall after every pass—that’s your secret to not circling forever.
• Obsess Over Boundaries:

  Especially when n is odd—the very center cell loves to be overcooked.

💻 The Code

public int[][] generateMatrix(int n) {
    int[][] matrix = new int[n][n];
    int left = 0, right = n - 1, top = 0, bottom = n - 1;
    int num = 1;
    while (num <= n * n) {
        for (int i = left; i <= right; i++) matrix[top][i] = num++;
        top++;
        for (int i = top; i <= bottom; i++) matrix[i][right] = num++;
        right--;
        for (int i = right; i >= left; i--) matrix[bottom][i] = num++;
        bottom--;
        for (int i = bottom; i >= top; i--) matrix[i][left] = num++;
        left++;
    }
    return matrix;
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Sneaky Edge Cases: n = 1 or 2? Those baby-sized matrices love sabotaging your beautiful bounds.
• Boundary Skipping: Update one bound too soon/late and your spiral looks like a Picasso on decaf.
• O(n²) for Both Time and Space. That’s as optimal as it gets—yes, even if your interviewer raises an eyebrow. Touch every cell, store every cell; your RAM will be fine.

🧠 Interviewer Brain-Tattoo

“Every spiral problem is just a boundary problem in a fancy suit. Turn the corners, shrink the suit, don’t double-tap the tie.”