local numbers = {3,6,4,2,1,5,2,5,5,5}
function sumwewant(array, sum)
local matches = {}
local alreadyMatched = {}
for i = 1, #array do
local first = array[i]
for j = 1, #array do
local second = numbers[j]
if i ~= j and not table.find(alreadyMatched, first) and not table.find(alreadyMatched, second) then
if ((first + second) == sum)then
table.insert(matches, {first, second})
table.insert(alreadyMatched, first)
table.insert(alreadyMatched, second)
end
end
end
end
return matches
end
local d = sumwewant(numbers, 10)
print(d)
local numbers = {3,6,4,5,2,1,5,2}
for i = 1, #numbers do
local first = numbers[i]
for j = i, #numbers do
local second = numbers[j]
if ((first + second) == 10)then
print(first, second)
break
end
end
Again I changed j = i
I just needed to make for each of j = i numbers. but anyways your script is still better and yea.
To be a little more exact, i should only count to the second to last element. j = i + 1 is so that the indices are not equal.
function sumwewant(array, sum)
local tab = {}
for i = 1, #array - 1 do
local first = array[i]
for j = i + 1, #array do
local second = numbers[j]
if ((first + second) == sum)then
table.insert(tab, {first, second})
end
end
end
return tab
end
But your method will work anyway because the second for loop will not execute that extra cycle because it will be out of range.
If you want it to have no duplicates just apply a filter before inserting a pair in the result.
local numbers = {3,6,4,2,1,5,2,5,5,5}
function sumwewant(array, sum)
local tab = {}
for i = 1, #array - 1 do
local first = array[i]
for j = i + 1, #array do
local second = numbers[j]
if ((first + second) == sum)then
local notInTab = true
for k = 1, #tab do
if tab[k][1] == first and tab[k][2] == second then
notInTab = false
break
end
end
if notInTab then
table.insert(tab, {first, second})
end
end
end
end
return tab
end
local d = sumwewant(numbers, 10)
print(d)
Actually it isnt a dublicate. It just print the Output {5,5} 2times so it doesnt mean there exist 2 dublicates because it just print the sameoutput 2 times. so there woudnt be a problem for my script and the things I want to create. Dublicate would be if it does print the output 2 times like 5,5 and then 5,5. I could easily check if it does print 1 times and the problem is solved.
This is what you actually mean!
This is what my script does, as you can see it print 2 times the same and it isnt a big problem,
I dont know why you post again something that was solved by @heII_ish but anyways my script is solved 2 times.
I got the impression that you wanted something more efficient, so I posted my answer. If you examine it closely you will notice that the changes, although subtle, are significant and, from my point of view, it is the right way to go.
