_885. Spiral Matrix III.md

885. Spiral Matrix III

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First completed : August 08, 2024

Last updated : August 08, 2024

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Related Topics : Array, Matrix, Simulation

Acceptance Rate : 84.66 %

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Notes

    Observing the samples, the distance of each "side" travelled follows
    the order of 1, 1, 2, 2, 3, 3, 4, 4, ... etc.

    I.e. we repeat a side length then increment.

    This makes sense as it follows whether we're traversing
    the height or width of the current "square."


    It is possible to further optimize this by "skipping" the
    out-of-bounds portions but this optimization isn't necessary for
    this medium question.
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Solutions

• m885 v1.py

Python

m885 v1.py

class Solution:
    def spiralMatrixIII(self, rows: int, cols: int, rStart: int, cStart: int) -> List[List[int]]:
        output = [[rStart, cStart]]

        distance = 1        # current side length
        direction = 0       # 0=r, 1=d, 2=l, 3=u
        twice = False

        while len(output) < rows * cols :
            for i in range(1, distance + 1) :
                match direction % 4 :
                    case 0 :
                        cStart += 1
                    case 1 :
                        rStart += 1
                    case 2 :
                        cStart -= 1
                    case 3 :
                        rStart -= 1

                if not (0 <= cStart < cols) or not (0 <= rStart < rows) :
                    continue

                output.append([rStart, cStart])

            direction += 1

            twice = not twice
            if not twice :
                distance += 1

        return output