caleb-bb
Optimizing an "almost-increasing" array problem
Below, the problem specs:
Given a sequence of integers as an array, determine whether it is possible to obtain a strictly increasing sequence by removing no more than one element from the array.
Note: sequence
a0,a1, …,anis considered to be a strictly increasing ifa0 < a1 < ... < an. Sequence containing only one element is also considered to be strictly increasing.Example
- For
sequence = [1, 3, 2, 1], the output should be
solution(sequence) = false.There is no one element in this array that can be removed in order to get a strictly increasing sequence.- For
sequence = [1, 3, 2], the output should be
solution(sequence) = true.You can remove3from the array to get the strictly increasing sequence[1, 2]. Alternately, you can remove2to get the strictly increasing sequence[1, 3].Input/Output
- [execution time limit] 12 seconds (exs)
- [input] array.integer sequenceGuaranteed constraints:
2 ≤ sequence.length ≤ 105,
-105 ≤ sequence[i] ≤ 105.- [output] booleanReturn
trueif it is possible to remove one element from the array in order to get a strictly increasing sequence, otherwise returnfalse.
Okay, so I wrote this to solve it:
def solution(sequence) do
iterate(sequence, 0)
end
def strictly_increasing?(list) do
list == Enum.uniq(list) |> Enum.sort
end
def iterate(enumerable, index) do
altered = List.delete_at(enumerable, index)
cond do
strictly_increasing?(altered) -> true
index == length(altered) -> false
true -> iterate(enumerable, index+1)
end
end
The problem with my solution is that it works, but it times out on the “submit” tests. Is there any way to speed this up, or do I need to switch to a totally difference algorithm?
Marked As Solved
al2o3cr
Ya know what else has to iterate over the whole list? List.delete_at, especially when called with length(list). A quick benchmark:
Mix.install([:benchee])
list1 = 1..100 |> Enum.to_list()
length1 = length(list1)
list2 = 1..1000 |> Enum.to_list()
length2 = length(list2)
list3 = 1..10000 |> Enum.to_list()
length3 = length(list3)
Benchee.run(
%{
"100 elements" => fn -> List.delete_at(list1, length1) end,
"1000 elements" => fn -> List.delete_at(list2, length2) end,
"10000 elements" => fn -> List.delete_at(list3, length3) end
},
time: 10,
memory_time: 2
)
produces the output on my machine:
Operating System: macOS
CPU Information: Intel(R) Core(TM) i7-6820HQ CPU @ 2.70GHz
Number of Available Cores: 8
Available memory: 16 GB
Elixir 1.13.4
Erlang 25.0.2
Benchmark suite executing with the following configuration:
warmup: 2 s
time: 10 s
memory time: 2 s
reduction time: 0 ns
parallel: 1
inputs: none specified
Estimated total run time: 42 s
Benchmarking 100 elements ...
Benchmarking 1000 elements ...
Benchmarking 10000 elements ...
Name ips average deviation median 99th %
100 elements 393.26 K 2.54 μs ±1944.85% 1.85 μs 4.95 μs
1000 elements 80.38 K 12.44 μs ±142.97% 10.82 μs 42.36 μs
10000 elements 6.85 K 145.95 μs ±33.72% 143.90 μs 279.72 μs
Comparison:
100 elements 393.26 K
1000 elements 80.38 K - 4.89x slower +9.90 μs
10000 elements 6.85 K - 57.40x slower +143.41 μs
Memory usage statistics:
Name Memory usage
100 elements 3.18 KB
1000 elements 20.95 KB - 6.59x memory usage +17.77 KB
10000 elements 224.88 KB - 70.72x memory usage +221.70 KB
Even calling length on the input is not desirable, since it has to traverse the list all the way to the end.
Calling List.delete_at as part of a recursion is a quick way to end up in the O(N^2) Bad Place.
A different way to think about this problem is to ask the question: what are the outcomes of adding an element to the end of an already almost-increasing list?
To express this with functions, we can use Enum.reduce_while/3 because of the control flow it gives us:
Enum.reduce_while(list, initial_state, fn element, state ->
# keep going:
{:cont, new_state}
# give up early:
{:halt, final_state}
end)
The trick with reduce and friends is to pick the right shape for the “state” data.
In this problem, we have two important parts of the state:
- the last element we want to compare the new one to
- what have we seen so far?
- almost-increasing with no errors
- almost-increasing with one error
- (no third case because we want to give up)
This suggests that “state” is a tuple of {status, last_element}.
Now we need an “initial” value. The empty list is almost-increasing, and we know the minimum value for elements is -105 so we can start with a value that will always be less than the first element of list:
Enum.reduce_while(list, {:all_increasing, -106}, fn element, {last_status, last_element} ->
...
end)
(Note: there’s nothing magical about :all_increasing, it’s chosen solely for human readability)
There are two factors that decide what to do next:
- the last state
- the comparison result between
last_elementandelement
so let’s use a case:
Enum.reduce_while(list, {:all_increasing, -106}, fn element, {last_status, last_element} ->
case {last_status, el > last_element} do
end
end)
The first case is easy: if the list is almost-increasing and el is bigger, then the list is still almost-increasing and we should next look at element:
Enum.reduce_while(list, {:all_increasing, -106}, fn element, {last_status, last_element} ->
case {last_status, element > last_element} do
{:all_increasing, true} ->
{:cont, {:all_increasing, element}}
end
end)
The next-easiest case is when we’ve already found a “bad” element and we find another one. In that case, it’s time to give up:
Enum.reduce_while(list, {:all_increasing, -106}, fn element, {last_status, last_element} ->
case {last_status, element > last_element} do
{:all_increasing, true} ->
{:cont, {:all_increasing, element}}
{:one_wrong, false} ->
{:halt, :not_increasing}
end
end)
Almost as easy: if we’ve seen one wrong element but the current one is increasing, keep going:
Enum.reduce_while(list, {:all_increasing, -106}, fn element, {last_status, last_element} ->
case {last_status, element > last_element} do
{:all_increasing, true} ->
{:cont, {:all_increasing, element}}
{:one_wrong, true} ->
{:cont, {:one_wrong, element}}
{:one_wrong, false} ->
{:halt, :not_increasing}
end
end)
The tricky one is if the current element is “bad” but there are no previous errors. In that case, we continue noting that there’s been an error AND skip el (since it’s bad):
Enum.reduce_while(list, {:all_increasing, -106}, fn element, {last_status, last_element} ->
case {last_status, element > last_element} do
{:all_increasing, true} ->
{:cont, {:all_increasing, element}}
{:all_increasing, false} ->
{:cont, {:one_wrong, last_element}}
{:one_wrong, true} ->
{:cont, {:one_wrong, element}}
{:one_wrong, false} ->
{:halt, :not_increasing}
end
end)
Without passing last_element in that case, sequences like [1,2,1,2] aren’t correctly detected.
Finally, the result of Enum.reduce_while is the last accumulator so it needs to be cleaned up into the boolean result.
Final code with some cleanup:
defmodule Sequence do
def almost_increasing(sequence) do
case Enum.reduce_while(sequence, {:all_increasing, -1000}, &do_almost_increasing/2) do
{:all_increasing, _} -> true
{:one_wrong, _} -> true
:not_increasing -> false
end
end
defp do_almost_increasing(el, {state, last_element}) do
case {state, el > last_element} do
{:all_increasing, true} ->
{:cont, {:all_increasing, el}}
{:all_increasing, false} ->
{:cont, {:one_wrong, last_element}}
{:one_wrong, true} ->
{:cont, {:one_wrong, el}}
{:one_wrong, false} ->
{:halt, :not_increasing}
end
end
end
Also Liked
Eiji
How about this one?
defmodule Example do
@remove_limit 1
# function head for a default argument
def sample(list, removed_so_far \\ 0)
# check if we have reached a remove limit
def sample(_list, removed_so_far) when removed_so_far > @remove_limit, do: false
# if a limit is not reached so far and a list have last item left
def sample([_head], _removed_so_far), do: true
# if a limit is not reached so far and a list is empty
def sample([], _removed_so_far), do: true
# if first and second item are not a part of an increasing list
# then function calls itself without a second argument
# if that would fail (return false) then function calls again itself
# but this time without a first argument instead
def sample([first, second | tail], removed_so_far) when first >= second do
removed_so_far = removed_so_far + 1
sample([first | tail], removed_so_far) || sample([second | tail], removed_so_far)
end
# in any other case we just skip a first item and perform check for a list's tail
def sample([_first, second | tail], removed_so_far) do
sample([second | tail], removed_so_far)
end
end
cmo
Try doing it with recursion, checking each pair (or the one before if you’re skipping this one), rather than checking the entire array every iteration.
Eiji
oh, my bad then
Can you please check the updated version?
defmodule Example do
@remove_limit 1
# function head for a default argument
def sample(list, previous \\ nil, removed_so_far \\ 0)
# check if we have reached a remove limit
def sample(_list, _previous, removed_so_far) when removed_so_far > @remove_limit, do: false
# if a limit is not reached so far and a list have last item left
def sample([_head], _previous, _removed_so_far), do: true
# if a limit is not reached so far and a list is empty
def sample([], _previous, _removed_so_far), do: true
# ensure nil which as atom is bigger than any integer is not part of check
def sample([nil | tail], previous, removed_so_far), do: sample(tail, previous, removed_so_far)
# if first and second item are not a part of an increasing list
# then function calls itself without a second argument
# if that would fail (return false) then function calls again itself
# but this time without a first argument instead
def sample([first, second | tail], previous, removed_so_far) when first >= second do
removed_so_far = removed_so_far + 1
sample([first | tail], previous, removed_so_far) ||
sample([previous, second | tail], previous, removed_so_far)
end
# in any other case we just skip a first item and perform check for a list's tail
def sample([first, second | tail], _previous, removed_so_far) do
sample([second | tail], first, removed_so_far)
end
end
When removing first from unmatched check I have added previous to list in order to perform check properly in that specific case.







