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KellenWatt's solution

to Spiral Matrix in the Go Track

Published at Apr 06 2020 · 0 comments
Instructions
Test suite
Solution

Given the size, return a square matrix of numbers in spiral order.

The matrix should be filled with natural numbers, starting from 1 in the top-left corner, increasing in an inward, clockwise spiral order, like these examples:

Spiral matrix of size 3
1 2 3
8 9 4
7 6 5
Spiral matrix of size 4
 1  2  3 4
12 13 14 5
11 16 15 6
10  9  8 7

Coding the solution

Look for a stub file having the name spiral_matrix.go and place your solution code in that file.

Running the tests

To run the tests run the command go test from within the exercise directory.

If the test suite contains benchmarks, you can run these with the --bench and --benchmem flags:

go test -v --bench . --benchmem

Keep in mind that each reviewer will run benchmarks on a different machine, with different specs, so the results from these benchmark tests may vary.

Further information

For more detailed information about the Go track, including how to get help if you're having trouble, please visit the exercism.io Go language page.

Source

Reddit r/dailyprogrammer challenge #320 [Easy] Spiral Ascension. https://www.reddit.com/r/dailyprogrammer/comments/6i60lr/20170619_challenge_320_easy_spiral_ascension/

Submitting Incomplete Solutions

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

spiral_matrix_test.go

package spiralmatrix

import (
	"reflect"
	"testing"
)

var testCases = []struct {
	description string
	input       int
	expected    [][]int
}{
	{
		description: "empty spiral",
		input:       0,
		expected:    [][]int{},
	},
	{
		description: "trivial spiral",
		input:       1,
		expected: [][]int{
			{1},
		},
	},
	{
		description: "spiral of size 2",
		input:       2,
		expected: [][]int{
			{1, 2},
			{4, 3},
		},
	},
	{
		description: "spiral of size 3",
		input:       3,
		expected: [][]int{
			{1, 2, 3},
			{8, 9, 4},
			{7, 6, 5},
		},
	},
	{
		description: "spiral of size 4",
		input:       4,
		expected: [][]int{
			{1, 2, 3, 4},
			{12, 13, 14, 5},
			{11, 16, 15, 6},
			{10, 9, 8, 7},
		},
	},
}

func TestSpiralMatrix(t *testing.T) {
	for _, testCase := range testCases {
		matrix := SpiralMatrix(testCase.input)
		if !reflect.DeepEqual(matrix, testCase.expected) {
			t.Fatalf("FAIL: %s\n\tSpiralMatrix(%v)\nexpected: %v\ngot     : %v",
				testCase.description, testCase.input, testCase.expected, matrix)
		}
		t.Logf("PASS: %s", testCase.description)
	}
}

func BenchmarkSpiralMatrix(b *testing.B) {
	for i := 0; i < b.N; i++ {
		for _, testCase := range testCases {
			SpiralMatrix(testCase.input)
		}
	}
}
package spiralmatrix

const (
	Up = iota
	Right
	Down
	Left
)


func SpiralMatrix(n int) [][]int {
	mat := make([][]int, n)
	for i:=0; i<n; i++ {
		mat[i] = make([]int, n)
	}

	p := [2]int{0,0}
	mod := [2]int{0,1}
	dir := Right

	for i:=1; i<=n*n; i++ {
		row,col := p[0],p[1]
		mat[row][col] = i
		if col+mod[1] >= n || col+mod[1] < 0 ||
		   row+mod[0] >= n || row+mod[0] < 0 ||
		   mat[row + mod[0]][col + mod[1]] != 0 {
			dir = (dir+1) % 4
			switch dir {
			case Up:
				mod = [2]int{-1,0}
			case Right:
				mod = [2]int{0,1}
			case Down:
				mod = [2]int{1,0}
			case Left:
				mod = [2]int{0,-1}
			}
		}
		p[0] += mod[0]
		p[1] += mod[1]
	}

	return mat
}

KellenWatt's Reflection

There's probably some sort of mathematical solution to this, but it didn't readily occur to me. Approaching this with a walker is easy enough, and it has relatively minimal overhead. Still O(n^2) complexity, just with a slightly larger coefficient than an equation.