Problem Description

You are given a grid with dimensions rows x cols and a starting position at cell (rStart, cStart). Your task is to traverse the entire grid in a clockwise spiral pattern, starting from the given position and initially facing east.

The traversal follows these rules:

• You move in a clockwise spiral: first east, then south, then west, then north, and repeat this pattern
• The spiral expands outward from your starting position
• When your path takes you outside the grid boundaries, you continue walking but only record positions that are within the valid grid (positions where 0 <= row < rows and 0 <= col < cols)
• You must visit every cell in the grid exactly once

The grid's coordinate system works as follows:

• The northwest (top-left) corner is at position (0, 0)
• The southeast (bottom-right) corner is at position (rows-1, cols-1)

Your goal is to return a list of coordinates [[r0, c0], [r1, c1], ...] representing all grid positions in the order they were visited during the spiral walk.

For example, if you start at the center of a grid and spiral outward, you would move:

1. One step east
2. One step south
3. Two steps west
4. Two steps north
5. Three steps east
6. Three steps south ... and so on, expanding the spiral with each complete rotation

The challenge is to correctly implement this expanding spiral pattern while keeping track of which positions fall within the grid boundaries.

Intuition

The key insight is recognizing the pattern of movement in a clockwise spiral. When we trace out a spiral path, we notice that we move in groups of steps, and each group follows a specific pattern:

• Move right for k steps
• Move down for k steps
• Move left for k+1 steps
• Move up for k+1 steps

After completing these four directions, we've made one complete "loop" of the spiral. The crucial observation is that after each complete loop, the number of steps increases by 2 for the next loop. This is because we need to cover a larger perimeter as the spiral expands outward.

Starting with k = 1:

• First loop: right 1, down 1, left 2, up 2
• Second loop: right 3, down 3, left 4, up 4
• Third loop: right 5, down 5, left 6, up 6
• And so on...

Why does this pattern work? Think about drawing a spiral on paper - after moving right and down by the same amount, you need to move one extra step left (to go past your starting column) and one extra step up (to go past your starting row) to complete the rectangle and position yourself for the next larger rectangle.

The direction changes can be represented as changes in row and column indices:

• Right: (0, 1) - row stays same, column increases
• Down: (1, 0) - row increases, column stays same
• Left: (0, -1) - row stays same, column decreases
• Up: (-1, 0) - row decreases, column stays same

Since we might walk outside the grid boundaries during our spiral, we only record positions when they satisfy 0 <= row < rows and 0 <= col < cols. We continue walking the spiral pattern regardless of whether we're inside or outside the grid, ensuring we eventually cover all cells within the grid.

The algorithm naturally terminates when we've collected exactly rows * cols valid positions, meaning we've visited every cell in the grid exactly once.

Solution Approach

We implement the spiral traversal using a systematic approach that tracks our current position and manages the expanding spiral pattern.

Initialization:

• Start with ans = [[rStart, cStart]] to record our starting position
• Handle the edge case where the grid has only one cell by returning immediately
• Initialize k = 1 to represent the number of steps in the first two directions of our spiral

Main Loop Structure: The algorithm uses an infinite while True loop that continues until we've visited all cells. Inside this loop, we iterate through four directions for each complete spiral rotation:

for dr, dc, dk in [[0, 1, k], [1, 0, k], [0, -1, k + 1], [-1, 0, k + 1]]:

This list encodes:

• [0, 1, k]: Move right - row change = 0, column change = 1, for k steps
• [1, 0, k]: Move down - row change = 1, column change = 0, for k steps
• [0, -1, k+1]: Move left - row change = 0, column change = -1, for k+1 steps
• [-1, 0, k+1]: Move up - row change = -1, column change = 0, for k+1 steps

Step-by-Step Movement: For each direction, we take the specified number of steps:

for _ in range(dk):
    rStart += dr
    cStart += dc

After each step, we check if the new position is within the grid boundaries:

if 0 <= rStart < rows and 0 <= cStart < cols:
    ans.append([rStart, cStart])

We only record positions that are valid grid cells. Positions outside the grid are skipped, but we continue walking the spiral pattern.

Termination Condition: After adding each valid position, we check if we've collected all cells:

if len(ans) == rows * cols:
    return ans

Once we've visited exactly rows * cols positions, we've covered the entire grid and can return the result.

Spiral Expansion: After completing each full rotation (all four directions), we increment k by 2:

k += 2

This ensures the spiral expands properly - the next rotation will have 2 more steps in each direction compared to the previous rotation.

The beauty of this approach is its simplicity - we don't need complex state management or direction tracking. The pattern [k, k, k+1, k+1] naturally creates the expanding spiral, and the boundary checking ensures we only record valid positions while maintaining the correct traversal order.

Example Walkthrough

Let's walk through a small example with a 3x3 grid starting at position (1, 1) (the center).

Grid:          Starting at (1,1):
[0,0] [0,1] [0,2]        . . .
[1,0] [1,1] [1,2]        . X .
[2,0] [2,1] [2,2]        . . .

Initial State:

• Position: (1, 1)
• ans = [[1, 1]] (starting position recorded)
• k = 1 (initial step count)

First Spiral Loop (k=1):

1. Move Right 1 step: (0, 1, k=1)

   • (1,1) → (1,2)
   • Valid position ✓, add [1,2]
   • ans = [[1,1], [1,2]]

2. Move Down 1 step: (1, 0, k=1)

   • (1,2) → (2,2)
   • Valid position ✓, add [2,2]
   • ans = [[1,1], [1,2], [2,2]]

3. Move Left 2 steps: (0, -1, k+1=2)

   • Step 1: (2,2) → (2,1)
     • Valid ✓, add [2,1]
   • Step 2: (2,1) → (2,0)
     • Valid ✓, add [2,0]
   • ans = [[1,1], [1,2], [2,2], [2,1], [2,0]]

4. Move Up 2 steps: (-1, 0, k+1=2)

   • Step 1: (2,0) → (1,0)
     • Valid ✓, add [1,0]
   • Step 2: (1,0) → (0,0)
     • Valid ✓, add [0,0]
   • ans = [[1,1], [1,2], [2,2], [2,1], [2,0], [1,0], [0,0]]

After first loop: k = 3 (incremented by 2)

Second Spiral Loop (k=3):

5. Move Right 3 steps: (0, 1, k=3)
   • Step 1: (0,0) → (0,1)
     • Valid ✓, add [0,1]
   • Step 2: (0,1) → (0,2)
     • Valid ✓, add [0,2]
   • ans = [[1,1], [1,2], [2,2], [2,1], [2,0], [1,0], [0,0], [0,1], [0,2]]
   Check: len(ans) = 9 = 3×3 ✓ All cells visited!

Final Path Visualization:

Order of visits:
  7   8   9        [0,0] [0,1] [0,2]
  6   1   2   →    [1,0] [1,1] [1,2]
  5   4   3        [2,0] [2,1] [2,2]

The spiral expands outward from the center, visiting all 9 cells in the correct clockwise pattern. Notice how the algorithm continues walking even when it would go outside the grid (which doesn't happen in this centered example), but only records valid positions.

Solution Implementation

Time and Space Complexity

Time Complexity: O(max(rows, cols)²)

The algorithm generates a spiral pattern starting from (rStart, cStart) and expanding outward. In each iteration of the outer while loop, the variable k increases by 2, and the spiral moves in 4 directions (right, down, left, up) with steps of k, k, k+1, k+1 respectively.

• The spiral continues until all rows * cols cells are visited
• In the worst case, the starting position is at a corner of the grid, requiring the spiral to expand to cover the entire grid
• The maximum distance from any starting point to cover all cells is proportional to max(rows, cols)
• For each complete square of the spiral, we visit O(k) cells where k goes from 1 to approximately 2 * max(rows, cols)
• Total iterations: 1 + 3 + 5 + 7 + ... + (2 * max(rows, cols) - 1) = O(max(rows, cols)²)

Even though we only store valid cells (those within bounds), we still need to traverse all positions in the spiral pattern, including those outside the grid boundaries.

Space Complexity: O(rows * cols)

The space complexity is determined by the output array ans which stores all valid positions within the grid. Since we store exactly rows * cols positions (each cell of the grid exactly once), the space complexity is O(rows * cols).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Spiral Step Pattern

One of the most common mistakes is getting the step count pattern wrong. Many developers incorrectly assume the spiral should use the same number of steps in all four directions, like [k, k, k, k], or might try [k, k+1, k, k+1].

Why it fails: The correct pattern is [k, k, k+1, k+1]. This is because after moving right and down by k steps, we need to move left and up by k+1 steps to complete the rectangular spiral layer. This ensures proper expansion of the spiral.

Solution:

# Correct pattern
directions = [
    (0, 1, steps_count),      # Right: k steps
    (1, 0, steps_count),      # Down: k steps  
    (0, -1, steps_count + 1), # Left: k+1 steps
    (-1, 0, steps_count + 1)  # Up: k+1 steps
]

2. Modifying Loop Variables Inside Nested Loops

Another pitfall is accidentally using the same variable names in nested loops or modifying the position variables incorrectly:

# WRONG - modifying rStart and cStart directly loses original reference
for dr, dc, dk in directions:
    for _ in range(dk):
        rStart += dr  # This modifies the parameter directly
        cStart += dc

Why it fails: While this works in Python (parameters are local variables), it makes the code harder to understand and maintain. Using descriptive variable names like current_row and current_col makes the intent clearer.

3. Off-by-One Error in Boundary Checking

A subtle mistake is using incorrect boundary conditions:

# WRONG - uses <= for upper bounds
if 0 <= current_row <= rows and 0 <= current_col <= cols:
    result.append([current_row, current_col])

Why it fails: Grid indices are 0-based, so valid row indices are from 0 to rows-1, not 0 to rows. The condition should be current_row < rows and current_col < cols.

4. Forgetting to Handle Single-Cell Grid

While not strictly necessary (the main loop will handle it), forgetting the optimization for a 1x1 grid can lead to unnecessary computation:

# Missing optimization
result = [[rStart, cStart]]
# Directly entering the while loop without checking rows * cols == 1

Why it's suboptimal: For a single-cell grid, we already have the complete answer after initialization. The early return if rows * cols == 1: return result avoids entering the loop unnecessarily.

5. Incorrect Spiral Expansion Rate

Some implementations forget to increment the step count properly:

# WRONG - incrementing by 1 instead of 2
steps_count += 1

# WRONG - forgetting to increment at all
# (no increment after completing 4 directions)

Why it fails: The spiral must expand by 2 steps after each complete rotation (all 4 directions). This is because after completing a full rectangular layer, the next layer needs one additional step on each side compared to the previous layer, resulting in a total increase of 2 steps per complete spiral.