List replication challenge with Erlang - erlang

Question

I'm trying to study Erlang resolving Hackerrank problems. There is a problem called List Replication. I finished with a solution like this:
-module(solution).
-export([main/0]).
process_input(Repeats)->
Line = io:get_line(""),
case string:len(Line) of
1 -> ok;
_ -> output(Line, Repeats), process_input(Repeats)
end.
output(_, 0)-> ok;
output(Line, Repeats)->
io:format(Line),
output(Line, Repeats - 1).
main()->
{ok, [Repeats]} = io:fread("", "~d"),
process_input(Repeats).
But this solution has a problem: I expect the last line is empty (in fact the last line is a number without \n). Any thoughts?

Answer 1

I think it's easier to handle eof from io:get_line/1 in case of a missing final newline, as well as handling a blank line to indicate end of input:
-module(solution).
-export([start/0]).
process_input(Repeat, Acc)->
case io:get_line("") of
Done when Done == eof; Done == "\n" ->
output(lists:reverse(Acc));
Line ->
Val = string:strip(Line, right, $\n),
Str = lists:duplicate(Repeat, Val),
process_input(Repeat, [Str|Acc])
end.
output(Lines)->
Out = [string:join(L, "\n")++"\n" || L <- Lines],
io:format("~s", [Out]).
start()->
{ok, [Repeat]} = io:fread("", "~d"),
process_input(Repeat, []).
The process_input/2 function now takes an accumulator, which is initially an empty list. It calls itself recursively until it detects end of input, after which it prints the output. It calls io:get_line/1 and checks to see if it returns either eof or just a newline, and for that case it reverses its accumulator and prints its output. For any other input, it strips the final newline, repeats the input via lists:duplicate/2, stores the result in a new accumulator, and passes that to a recursive call.
The output/1 function takes the accumulator from process_input/2, joins the repeated values with newlines, then prints the results. Note that this version of the solution module restricts the formatting of the result to the output/1 function, in case you wanted to use the raw result of process_input/2 for some other purpose.
And finally, I renamed your main/0 to start/0 because running a function via the erl -s command line option assumes a function named start if none is given.
We can use printf in a Unix shell to create an input file without a final newline:
$ printf '3\n1\n2\n3\n4\n' > in
and then run our compiled solution module like this, getting the tripled output we expect in this case:
$ cat in | erl -noshell -s solution -s init stop
1
1
1
2
2
2
3
3
3
4
4
4
Adding a final newline to the in file gives the same result (try it). We can also create input with more than a single character per line:
$ printf '2\nhey\nhey\nwhat\ncan\nI\ndo' > in2
$ cat in2 | erl -noshell -s solution -s init stop
hey
hey
hey
hey
what
what
can
can
I
I
do
do
and for this in2 file, we get the doubled output we expect as well.

Answer 2

You will have to turn off the echo of the stream, with io:setopts/1
-module(solution).
-export([main/0]).
process_input(Repeats)->
Line = string:strip(io:get_line(""), right, $\n),
case string:len(Line) of
0 ->
ok;
_ ->
output(Line, Repeats),
process_input(Repeats)
end.
output(_, 0)->
ok;
output(Line, Repeats)->
io:format("~s\n", [Line]),
output(Line, Repeats - 1).
main()->
{ok, [Repeats]} = io:fread("", "~d"),
io:setopts([{echo, false}]),
process_input(Repeats).

Answer 3

I just updated the code which solves the challenge, so the idea is that you save the input in an accumulator (a list in this case) and at the end you process the list, if you want to avoid the output of what is being written on screen at any moment you can disable the echo as the example I show on the other answer
-module(solution).
-export([main/0]).
process_input(Data)->
Number = io:get_line(""),
case string:len(Number) of
1 ->
Data;
_ ->
process_input(Data ++ [Number])
end.
process_output([], _)->
ok;
process_output([H|Data], Repeats)->
print(H, Repeats),
process_output(Data, Repeats).
print(_, 0) ->
ok;
print(Element, Times) ->
io:format(Element),
print(Element, Times - 1).
main()->
{ok, [Repeats]} = io:fread("", "~d"),
Data = process_input([]),
process_output(Data, Repeats).
And the test:
rorra:~/erlang > erl
Erlang/OTP 17 [erts-6.4] [source] [64-bit] [smp:8:8] [async-threads:10] [hipe] [kernel-poll:false]
Eshell V6.4 (abort with ^G)
1> c(solution).
{ok,solution}
2> solution:main().
3
1
2
3
4
1
1
1
2
2
2
3
3
3
4
4
4
ok
3>

Answer 4

My own answer:
-module(solution).
-export([main/0]).
process_input(Repeats)->
Line = io:get_line(""),
case lists:reverse(Line) of
"\n" ++ _ -> output(Line, Repeats), process_input(Repeats);
_ -> output(Line ++ "~n" , Repeats)
end.
output(_, 0)-> ok;
output(Line, Repeats)->
io:format(Line),
output(Line, Repeats - 1).
main()->
{ok, [Repeats]} = io:fread("", "~d"),
process_input(Repeats).

Answer 5

Here's what I came up with (with help from the answers above). I'm adding my solution because it also handles the constraints stated in the problem:
-module(solution).
-export([main/0]).
main() -> get_input([]).
print_list(_,[_|[]]) -> ok;
print_list(NumTimes,[_|List]) ->
Number = hd(List),
print_list_number(NumTimes,Number),
print_list(NumTimes,List).
print_list_number(NumTimes,Number)
when
(NumTimes > 0)
and
(Number >= 1)
and
(Number =< 100)
->
io:fwrite("~B~n",[Number]),
print_list_number(NumTimes - 1,Number);
print_list_number(_,_) -> ok.
get_input(Acc) ->
case io:get_line("") of
Done when
Done == eof;
Done == "\n"
->
NumTimes = hd(Acc),
%% Make sure we only run this when the List length isn't greater than 10
if
(length(Acc) - 1 =< 10)
and
(NumTimes =< 100)
->
print_list(NumTimes,Acc);
true -> ok
end;
Line ->
{NumInt,_} = string:to_integer(Line),
List = Acc ++ [NumInt],
get_input(List)
end.
I'm pretty new to Erlang so forgive me if this is messy / inefficient!