Lua Compare two Tables values No Order - performance

Question

I wonder if there is a faster way to check if two tables are equal (without order in the keys).
This is my solution
function TableCompareNoOrder(table1,table2)
if #table1 ~= #table2 then return false end
local equal = false
for key, item in pairs(table1) do
for key2, item2 in pairs(table2) do
if item == item2 then equal = true end
end
if equal == false then return false end
end
local equal = false
for key, item in pairs(table2) do
for key2, item2 in pairs(table1) do
if item == item2 then equal = true end
end
if equal == false then return false end
end
return equal
end
a = {1,2,3,0}
b = {2,3,1,0}
print(TableCompareNoOrder(a,b))

Answer 1

Yes, there are faster ways. Your approach is currently O(n²) as it has to compare each element with each other element. One way is to sort the arrays first. Another one would be to build a lookup table ("set") using the hash part. If hash access is assumed to take amortized constant time, this should run in O(n) time - significantly faster. It will take O(n) additional memory but leaves the passed tables intact; to do the same using sorting you'd have to first create a copy, then sort it.
The hash approach looks like the following (and also works for values like tables which are compared by reference and can't be sorted unless a metatable is set or a custom comparator function is supplied):
function TableCompareNoOrder(table1, table2)
if #table1 ~= #table2 then return false end
-- Consider an early "return true" if table1 == table2 here
local t1_counts = {}
-- Check if the same elements occur the same number of times
for _, v1 in ipairs(table1) do
t1_counts[v1] = (t1_counts[v1] or 0) + 1
end
for _, v2 in ipairs(table2) do
local count = t1_counts[v2] or 0
if count == 0 then return false end
t1_counts[v2] = count - 1
end
return true
end
For the sake of completeness, here's a simple implementation using sorting:
function TableCompareNoOrder(table1, table2)
if #table1 ~= #table2 then return false end
-- Lazy implementation: Sort copies of both tables instead of using a binary search. Takes twice as much memory.
local t1_sorted = {table.unpack(table1)} -- simple way to copy the table, limited by stack size
table.sort(t1_sorted)
local t2_sorted = {table.unpack(table2)}
table.sort(t2_sorted)
for i, v1 in ipairs(t1_sorted) do
if t2_sorted[i] ~= v1 then return false end
end
return true
end
This should roughly run in O(n log n) (performance is determined by the sorting algorithm, usually quicksort which has this as it's average running time).