Avatar of JaeHyoLee

JaeHyoLee's solution

to Sublist in the Lua Track

Published at Jul 13 2018 · 3 comments
Instructions
Test suite
Solution

Given two lists determine if the first list is contained within the second list, if the second list is contained within the first list, if both lists are contained within each other or if none of these are true.

Specifically, a list A is a sublist of list B if by dropping 0 or more elements from the front of B and 0 or more elements from the back of B you get a list that's completely equal to A.

Examples:

  • A = [1, 2, 3], B = [1, 2, 3, 4, 5], A is a sublist of B
  • A = [3, 4, 5], B = [1, 2, 3, 4, 5], A is a sublist of B
  • A = [3, 4], B = [1, 2, 3, 4, 5], A is a sublist of B
  • A = [1, 2, 3], B = [1, 2, 3], A is equal to B
  • A = [1, 2, 3, 4, 5], B = [2, 3, 4], A is a superlist of B
  • A = [1, 2, 4], B = [1, 2, 3, 4, 5], A is not a superlist of, sublist of or equal to B

Running the tests

To run the tests, run the command busted from within the exercise directory.

Further information

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

Submitting Incomplete Solutions

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

sublist_spec.lua

local is_sublist = require('sublist')

describe('sublist', function()
  it('should consider an empty list to be a sublist of an empty list', function()
    assert.equal(true, is_sublist({}, {}))
  end)

  it('should consider an empty list to be a sublist of a non-empty list', function()
    assert.equal(true, is_sublist({}, { 1, 2, 3 }))
  end)

  it('should consider a list to be a sublist of itself', function()
    assert.equal(true, is_sublist({ 1, 2, 3 }, { 1, 2, 3 }))
  end)

  it('should not consider a subset to be a sublist', function()
    assert.equal(false, is_sublist({ 1, 2, 3 }, { 2, 1, 3 }))
  end)

  it('should find a sublist at the beginning of a list', function()
    assert.equal(true, is_sublist({ 11, 22, 33 }, { 11, 22, 33, 44, 55 }))
  end)

  it('should find a sublist in the middle of a list', function()
    assert.equal(true, is_sublist({ 12, 13, 14 }, { 11, 12, 13, 14, 15 }))
  end)

  it('should find a sublist at the end of a list', function()
    assert.equal(true, is_sublist({ 30, 40, 50 }, { 10, 20, 30, 40, 50 }))
  end)

  it('should be able to determine when a list is not a sublist', function()
    assert.equal(false, is_sublist({ 1, 2, 3 }, { 5, 6, 7, 8, 9 }))
  end)

  it('should not consider almost sublists to be sublists', function()
    assert.equal(false, is_sublist({ 3, 4, 5 }, { 1, 2, 4, 5, 6 }))
    assert.equal(false, is_sublist({ 3, 4, 5 }, { 1, 2, 3, 4, 6 }))
  end)

  it('should find a sublist when there are multiple instances of the sublist', function()
    assert.equal(true, is_sublist({ 1, 2, 3 }, { 0, 1, 2, 3, 4, 1, 2, 3, 6 }))
  end)
end)
local function make_set(a)
  local set = {}
  for _, l in ipairs(a) do
    set[l] = true
  end
  return set
end

return function (a, b)
  local c = make_set(b)
  for _, v in pairs(a) do
    if c[v] == nil then return false end
  end
  return true
end

Community comments

Find this solution interesting? Ask the author a question to learn more.
Avatar of ryanplusplus

Great use of Lua tables to minimize search time :)

Avatar of fyrchik

You are finding a subset here. But {1,2} is not sublist of {2,1}!

Avatar of ryanplusplus

@fyrchik you're absolutely right, I didn't notice that before. Looks like this problem is missing a test.

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