I'm doing some codewars and arr[index] keeps returning nil. I've done this a few different ways, and I'm sure the array exists, as well as the index. What's wrong here, is it syntax?
As I've mentioned in the title, I want to find the last digit of the array.
if arr[index] <= 0 then
return -1
end
Full Code:
local solution = {}
function solution.newAvg(arr, navg)
local currentAverage = 0
local index = 0
for i, v in pairs(arr) do
index = i
currentAverage = currentAverage + v
end
if arr[index] <= 0 then
return -1
end
return math.ceil(((index+1) * navg) - currentAverage)
end
return solution
I see two issues with your code:
Edge case: Empty array
If arr = {}, the loop for i, v in pairs(arr) do won't execute at all and index will remain at 0. Since arr is empty, arr[0] will be nil and arr[index] <= 0 will fail with an "attempt to compare a nil value" error.
Lack of ordering guarantee
You use pairs rather than ipairs to loop over what I assume is a list. This means keys & values might be traversed in any order. In practice pairs usually (but not always!) traverses the list part of a table in the same order as ipairs, but the reference manual clearly states that you can't rely on no specific order. I don't think CodeWars is this advanced but consider the possibility that pairs may be overridden to deliberately shuffle the order of traversal in order to check whether you're relying on the dreaded "undefined behavior". If this is the case, your "last index" might actually be any index that happens to be visited last, obviously breaking your algorithm.
Fixes
I'll assume arr is an "array", that is, it only contains keys from 1 to n and all values are non-nil (i.e. there are no holes). Then you can (and should!) use ipairs to loop over the "array":
for i, v in ipairs(arr) do ... end
I don't know the problem statement so it's hard to tell how an empty array should be handled. I'll assume that it should probably return 0. You could add a simply early return at the top of the function for that: if arr[1] == nil then return 0 end. Nonempty arrays will always have arr[1] ~= nil.
I want to find the last digit of the array.
If you mean the last integer (or entry/item) of the array:
local last = array[#array]
If you mean the last digit (for example array = {10, 75, 44, 62} and you want 2), then you can get the last item and then get the last digit using modulo 10:
local last = array[#array] % 10
for i, v in pairs(arr) do
index = i
currentAverage = currentAverage + v
end
Just a reminder:
#array returns the number of items in a table.
In Lua, arrays are implemented using integer-indexed tables.
There's a difference between pairs() and ipairs().
Regarding point 3 above, the following code:
local array = {
[1] = 12,
[2] = 32,
[3] = 41,
[4] = 30,
[5] = 14,
[6] = 50,
[7] = 62,
[8] = 57
}
for key, value in pairs(array) do
print(key, value)
end
produces the following output (note that the order of keys is not respected):
8 57
1 12
2 32
3 41
4 30
5 14
6 50
7 62
while the same code above with pairs() replaced with ipairs() gives:
1 12
2 32
3 41
4 30
5 14
6 50
7 62
8 57
So, this might be the cause of your problem.
Related
I'm trying to create a code that generates random numbers within the range 10-30 but making sure that no number is repeated. It shows "subscript out of range" on NumArray(Count) = Count when I run the code.
'Make an array of completely sorted numbers
FOR Count = 10 TO 30
NumArray(Count) = Count
NEXT Count
RANDOMIZE TIMER
FOR Count = 10 TO 30
Number = (RND * (31 - Count)) + 10
PRINT #1, NumArray(Number)
FOR Counter = Number TO 30 - Count
NumArray(Counter) = NumArray(Counter + 1)
NEXT Counter
NEXT Count
This isn't actually my code. Copied and pasted for my assignment.
It looks like you're missing some DIM statements.
Variables containing numbers have type SINGLE by default, so you might see something like FOR Counter = 18.726493 TO 20 because the RND function returns a number between 0 and 1, excluding 1, meaning you will be trying to use NumArray(18.726493) which will not work.
Arrays that are not explicitly declared can only have 11 items with an index from 0 to 10, but the range 10-30 requires you to store 21 items (30 - 10 + 1 = 21). You can also specify a custom upper and lower bound if it will make your code easier for you to understand. Add these lines before the first line in your code shown above:
DIM Number AS INTEGER
DIM NumArray(10 TO 30) AS INTEGER
This will ensure Number only contains integers (any fractional values are rounded to the nearest integer), and NumArray will work from NumArray(10) to NumArray(30), but you can't use NumArray(9), NumArray(8), NumArray(31), etc. The index must be in the range 10-30.
I think that should fix your code, but I don't know for certain since I don't fully understand how it is supposed to work. At the very least, it will fix the type and subscript problems in your code.
You need to declare the array:
'Make an array of completely sorted numbers
DIM NumArray(30) AS INTEGER
FOR Count = 10 TO 30
NumArray(Count) = Count
NEXT Count
RANDOMIZE TIMER
FOR Count = 10 TO 30
Number = (RND * (31 - Count)) + 10
PRINT #1, NumArray(Number)
FOR Counter = Number TO 30 - Count
NumArray(Counter) = NumArray(Counter + 1)
NEXT Counter
NEXT Count
This question already has answers here:
Finding all possible combinations of numbers to reach a given sum
(32 answers)
Closed 6 years ago.
Need to create an array whose sum should be equal to expected value.
inp = [1,2,3,4,5,6,7,8,9,10]
sum = 200
output:
out = [10,10,9,1,3,3,3,7,.....] whose sum should be 200
or
out = [10,7,3,....] Repeated values can be used
or
out = [2,3,4,9,2,....]
I tried as,
arr = [5,10,15,20,30]
ee = []
max = 200
while (ee.sum < max) do
ee << arr.sample(1).first
end
ee.pop(2)
val = max - ee.sum
pair = arr.uniq.combination(2).detect { |a, b| a + b == val }
ee << pair
ee.flatten
Is there any effective way to do it.
inp = [1,2,3,4,5,6,7,8,9,10]
sum = 20
inp.length.downto(1).flat_map do |i|
inp.combination(i).to_a # take all subarrays of length `i`
end.select do |a|
a.inject(:+) == sum # select only those summing to `sum`
end
One might take a random element of resulting array.
result = inp.length.downto(1).flat_map do |i|
inp.combination(i).to_a # take all subarrays of length `i`
end.select do |a|
a.inject(:+) == sum # select only those summing to `sum`
end
puts result.length
#⇒ 31
puts result.sample
#⇒ [2, 4, 5, 9]
puts result.sample
#⇒ [1, 2, 3, 6, 8]
...
Please note, that this approach is not efficient for long-length inputs. As well, if any original array’s member might be taken many times, combination above should be changed to permutation, but this solution is too ineffective to be used with permutation.
I found an answer of this question in the following link:
Finding all possible combinations of numbers to reach a given sum
def subset_sum(numbers, target, partial=[])
s = partial.inject 0, :+
#check if the partial sum is equals to target
puts "sum(#{partial})=#{target}" if s == target
return if s >= target #if we reach the number why bother to continue
(0..(numbers.length - 1)).each do |i|
n = numbers[i]
remaining = numbers.drop(i+1)
subset_sum(remaining, target, partial + [n])
end
end
subset_sum([1,2,3,4,5,6,7,8,9,10],20)
The following script finds prime numbers in a range from 1 to 13.
When I explicitly iterate over the table that contains the results I can see that the script works as expected. However, if I use unpack() function on the table only the first 3 numbers get printed out.
From docs: unpack is "a special function with multiple returns. It receives an array and returns as results all elements from the array, starting from index 1".
Why is it not working in the script below?
t = {}
for i=1, 13 do t[i] = i end
primes = {}
for idx, n in ipairs(t) do
local isprime = true
for i=2, n-1 do
if n%i == 0 then
isprime = false
break
end
end
if isprime then
primes[idx] = n
end
end
print('loop printing:')
for i in pairs(primes) do
print(i)
end
print('unpack:')
print(unpack(primes))
Running
$ lua5.3 primes.lua
loop printing:
1
2
3
5
7
13
11
unpack:
1 2 3
Change
primes[idx] = n
to
primes[#primes+1] = n
The reason is that idx is not sequential as not every number is a prime.
I'm currently practicing programming problems and out of interest, I'm trying a few Codility exercises in Lua. I've been stuck on the Passing Cars problem for a while.
Problem:
A non-empty zero-indexed array A consisting of N integers is given. The consecutive elements of array A represent consecutive cars on a road.
Array A contains only 0s and/or 1s:
0 represents a car traveling east,
1 represents a car traveling west.
The goal is to count passing cars. We say that a pair of cars (P, Q), where 0 ≤ P < Q < N, is passing when P is traveling to the east and Q is traveling to the west.
For example, consider array A such that:
A[0] = 0
A[1] = 1
A[2] = 0
A[3] = 1
A[4] = 1
We have five pairs of passing cars: (0, 1), (0, 3), (0, 4), (2, 3), (2, 4).
Write a function:
function solution(A)
that, given a non-empty zero-indexed array A of N integers, returns the number of pairs of passing cars.
The function should return −1 if the number of pairs of passing cars exceeds 1,000,000,000.
For example, given:
A[0] = 0
A[1] = 1
A[2] = 0
A[3] = 1
A[4] = 1
the function should return 5, as explained above.
Assume that:
N is an integer within the range [1..100,000];
each element of array A is an integer that can have one of the following values: 0, 1.
Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
My attempt in Lua keeps failing but I can't seem to find the issue.
local function solution(A)
local zeroes = 0
local pairs = 0
for i = 1, #A do
if A[i] == 0 then
zeroes = zeroes + 1
else
pairs = pairs + zeroes
if pairs > 1e9 then
return -1
end
end
end
return pairs
end
In terms of time-space complexity constraints, I think it should pass so I can't seem to find the issue. What am I doing wrong? Any advice or tips to make my code more efficient would be appreciated.
FYI: I keep getting a result of 2 when the desired example result is 5.
The problem statement says A is 0-based so if we ignore the first and start at 1, the output would be 2 instead of 5. 0-based tables should be avoided in Lua, they go against convention and will lead to a lot of off-by one errors: for i=1,#A do will not do what you want.
function solution1based(A)
local zeroes = 0
local pairs = 0
for i = 1, #A do
if A[i] == 0 then
zeroes = zeroes + 1
else
pairs = pairs + zeroes
if pairs > 1e9 then
return -1
end
end
end
return pairs
end
print(solution1based{0, 1, 0, 1, 1}) -- prints 5 as you wanted
function solution0based(A)
local zeroes = 0
local pairs = 0
for i = 0, #A do
if A[i] == 0 then
zeroes = zeroes + 1
else
pairs = pairs + zeroes
if pairs > 1e9 then
return -1
end
end
end
return pairs
end
print(solution0based{[0]=0, [1]=1, [2]=0, [3]=1, [4]=1}) -- prints 5
myTable = {}
myTable["foo"] = 12
myTable["bar"] = "blah"
print(#myTable) -- this prints 0
Do I actually have to iterate through the items in the table to get the number of keys?
numItems = 0
for k,v in pairs(myTable) do
numItems = numItems + 1
end
print(numItems) -- this prints 2
I experimented with both the # operator and table.getn(). I thought table.getn() would do what you wanted but as it turns out it's returning the same value as #, namely 0. It appears that dictionaries insert nil placeholders as necessary.
Looping over the keys and counting them seems like the only way to get the dictionary size.
The length operator:
The length of a table t is defined to be any integer index n such that t[n] is not nil and t[n+1] is nil; moreover, if t[1] is nil, n can be zero. For a regular array, with non-nil values from 1 to a given n, its length is exactly that n, the index of its last value. If the array has "holes" (that is, nil values between other non-nil values), then #t can be any of the indices that directly precedes a nil value (that is, it may consider any such nil value as the end of the array).
so only way to get length is iterate over it.
Aside from iterating through the keys manually, it is simple to automatically keep track of it via metamethods. Considering you probably don't want to keep track of every table you make, you can just write a function that will allow you to convert any table into a key-countable object. The following isn't perfect, but I think it would illustrate the point:
function CountedTable(x)
assert(type(x) == 'table', 'bad parameter #1: must be table')
local mt = {}
-- `keys` will represent the number of non integral indexes
-- `indxs` will represent the number of integral indexes
-- `all` will represent the number of both
local keys, indxs, all = 0, 0, 0
-- Do an initial count of current assets in table.
for k, v in pairs(x) do
if (type(k) == 'number') and (k == math.floor(k)) then indxs = indxs + 1
else keys = keys + 1 end
all = all + 1
end
-- By using `__nexindex`, any time a new key is added, it will automatically be
-- tracked.
mt.__newindex = function(t, k, v)
if (type(k) == 'number') and (k == math.floor(k)) then indxs = indxs + 1
else keys = keys + 1 end
all = all + 1
t[k] = v
end
-- This allows us to have fields to access these datacounts, but won't count as
-- actual keys or indexes.
mt.__index = function(t, k)
if k == 'keyCount' then return keys
elseif k == 'indexCount' then return indxs
elseif k == 'totalCount' then return all end
end
return setmetatable(x, mt)
end
Examples of using this would include:
-- Note `36.35433` would NOT be counted as an integral index.
local foo = CountedTable { 1, 2, 3, 4, [36.35433] = 36.35433, [54] = 54 }
local bar = CountedTable { x = 23, y = 43, z = 334, [true] = true }
local foobar = CountedTable { 1, 2, 3, x = 'x', [true] = true, [64] = 64 }
print(foo.indexCount) --> 5
print(bar.keyCount) --> 4
print(foobar.totalCount) --> 6
Live Working Example
Hope this helped! :)
Lua stores table as two separates parts : a hash part and an array part, the len operator only deal with the array part, meaning value indexed by a number value, plus using rules mentioned below, so you don't have any choice for counting "hash" value you need to iterate over the table with pairs() function.