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j0ran's solution

to Spiral Matrix in the Go Track

Published at Aug 27 2019 · 1 comment
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

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

// SpiralMatrix returns a spiral matrix of size n
func SpiralMatrix(n int) [][]int {
	m := make([][]int, n)
	for y := 0; y < n; y++ {
		m[y] = make([]int, n)
	}

	x, y := 0, 0
	dx, dy := 1, 0
	for count := 1; count <= n*n; count++ {
		m[y][x] = count
		xx, yy := x+dx, y+dy
		if xx < 0 || xx >= n || yy < 0 || yy >= n || m[yy][xx] != 0 {
			dx, dy = -dy, dx
		}
		x, y = x+dx, y+dy
	}

	return m
}

Community comments

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Avatar of j0ran
Solution Author
commented 271 days ago

Looking at some other solutions I saw much better approach. I choose to try one of those but include my dx ,dy solution.

j0ran's Reflection

It's a non standard solution I think.