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:
To run the tests, run the command busted
from within the exercise directory.
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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
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Community comments
Great use of Lua tables to minimize search time :)
You are finding a subset here. But {1,2} is not sublist of {2,1}!
@fyrchik you're absolutely right, I didn't notice that before. Looks like this problem is missing a test.