Adjacent number algorithm grouper - algorithm

Question

By which I mean this:
Given the input set of numbers:
1,2,3,4,5 becomes "1-5".
1,2,3,5,7,9,10,11,12,14 becomes "1-3, 5, 7, 9-12, 14"
This is the best I managed to come up with: [C#]
Which feels a little sloppy to me, so the question is, is there somehow more readable and/or elegant solution to this?
public static string[] FormatInts(int[] ints)
{
if (ints == null)
throw new ArgumentNullException("ints"); // hey what are you doing?
if (ints.Length == 0)
return new string[] { "" }; // nothing to process
if (ints.Length == 1)
return new string[] { ints[0].ToString() }; // nothing to process
Array.Sort<int>(ints); // need to sort these lil' babies
List<string> values = new List<string>();
int lastNumber = ints[0]; // start with the first number
int firstNumber = ints[0]; // same as above
for (int i = 1; i < ints.Length; i++)
{
int current = ints[i];
int difference = (lastNumber - current ); // compute difference between last number and current number
if (difference == -1) // the numbers are adjacent
{
if (firstNumber == 0) // this is the first of the adjacent numbers
{
firstNumber = lastNumber;
}
else // we're somehow in the middle or at the end of the adjacent number set
{
lastNumber = current;
continue;
}
}
else
{
if (firstNumber > 0 && firstNumber != lastNumber) // get ready to print a set of numbers
{
values.Add(string.Format("{0}-{1}", firstNumber, lastNumber));
firstNumber = 0; // reset
}
else // print a single value
{
values.Add(string.Format("{0}", lastNumber));
}
}
lastNumber = current;
}
if (firstNumber > 0) // if theres anything left, print it out
{
values.Add(string.Format("{0}-{1}", firstNumber, lastNumber));
}
return values.ToArray();
}

Answer 1

I've rewritten your code like this:
public static string[] FormatInts(int[] ints)
{
Array.Sort<int>(ints);
List<string> values = new List<string>();
for (int i = 0; i < ints.Length; i++)
{
int groupStart = ints[i];
int groupEnd = groupStart;
while (i < ints.Length - 1 && ints[i] - ints[i + 1] == -1)
{
groupEnd = ints[i + 1];
i++;
}
values.Add(string.Format(groupEnd == groupStart ? "{0}":"{0} - {1}", groupStart, groupEnd));
}
return values.ToArray();
}
And then:
/////////////////
int[] myInts = { 1,2,3,5,7,9,10,11,12,14 };
string[] result = FormatInts(myInts); // now result haves "1-3", "5", "7", "9-12", "14"

Answer 2

See How would you display an array of integers as a set of ranges? (algorithm)
My answer to the above question:
void ranges(int n; int a[n], int n)
{
qsort(a, n, sizeof(*a), intcmp);
for (int i = 0; i < n; ++i) {
const int start = i;
while(i < n-1 and a[i] >= a[i+1]-1)
++i;
printf("%d", a[start]);
if (a[start] != a[i])
printf("-%d", a[i]);
if (i < n-1)
printf(",");
}
printf("\n");
}

Answer 3

Pure functional Python:
#!/bin/env python
def group(nums):
def collect((acc, i_s, i_e), n):
if n == i_e + 1: return acc, i_s, n
return acc + ["%d"%i_s + ("-%d"%i_e)*(i_s!=i_e)], n, n
s = sorted(nums)+[None]
acc, _, __ = reduce(collect, s[1:], ([], s[0], s[0]))
return ", ".join(acc)
assert group([1,2,3,5,7,9,10,11,12,14]) == "1-3, 5, 7, 9-12, 14"

Answer 4

I'm a bit late to the party, but anyway, here is my version using Linq:
public static string[] FormatInts(IEnumerable<int> ints)
{
var intGroups = ints
.OrderBy(i => i)
.Aggregate(new List<List<int>>(), (acc, i) =>
{
if (acc.Count > 0 && acc.Last().Last() == i - 1) acc.Last().Add(i);
else acc.Add(new List<int> { i });
return acc;
});
return intGroups
.Select(g => g.First().ToString() + (g.Count == 1 ? "" : "-" + g.Last().ToString()))
.ToArray();
}

Answer 5

Looks clear and straightforward to me. You can simplify a bit if you either assume the input array is sorted, or sort it yourself before further processing.
The only tweak I'd suggest would be to reverse the subtraction:
int difference = (current - lastNumber);
... simply because I find it easier to work with positive differences. But your code is a pleasure to read!

Answer 6

As I wrote in comment, I am not fan of the use of value 0 as flag, making firstNumber both a value and a flag.
I did a quick implementation of the algorithm in Java, boldly skipping the validity tests you already correctly covered...
public class IntListToRanges
{
// Assumes all numbers are above 0
public static String[] MakeRanges(int[] numbers)
{
ArrayList<String> ranges = new ArrayList<String>();
Arrays.sort(numbers);
int rangeStart = 0;
boolean bInRange = false;
for (int i = 1; i <= numbers.length; i++)
{
if (i < numbers.length && numbers[i] - numbers[i - 1] == 1)
{
if (!bInRange)
{
rangeStart = numbers[i - 1];
bInRange = true;
}
}
else
{
if (bInRange)
{
ranges.add(rangeStart + "-" + numbers[i - 1]);
bInRange = false;
}
else
{
ranges.add(String.valueOf(numbers[i - 1]));
}
}
}
return ranges.toArray(new String[ranges.size()]);
}
public static void ShowRanges(String[] ranges)
{
for (String range : ranges)
{
System.out.print(range + ","); // Inelegant but quickly coded...
}
System.out.println();
}
/**
* #param args
*/
public static void main(String[] args)
{
int[] an1 = { 1,2,3,5,7,9,10,11,12,14,15,16,22,23,27 };
int[] an2 = { 1,2 };
int[] an3 = { 1,3,5,7,8,9,11,12,13,14,15 };
ShowRanges(MakeRanges(an1));
ShowRanges(MakeRanges(an2));
ShowRanges(MakeRanges(an3));
int L = 100;
int[] anr = new int[L];
for (int i = 0, c = 1; i < L; i++)
{
int incr = Math.random() > 0.2 ? 1 : (int) Math.random() * 3 + 2;
c += incr;
anr[i] = c;
}
ShowRanges(MakeRanges(anr));
}
}
I won't say it is more elegant/efficient than your algorithm, of course... Just something different.
Note that 1,5,6,9 can be written either 1,5-6,9 or 1,5,6,9, not sure what is better (if any).
I remember having done something similar (in C) to group message numbers to Imap ranges, as it is more efficient. A useful algorithm.

Answer 7

Perl
With input validation/pre-sorting
You can easily get the result as a LoL if you need to do something more fancy than
just return a string.
#!/usr/bin/perl -w
use strict;
use warnings;
use Scalar::Util qw/looks_like_number/;
sub adjacenify {
my #input = #_;
# Validate and sort
looks_like_number $_ or
die "Saw '$_' which doesn't look like a number" for #input;
#input = sort { $a <=> $b } #input;
my (#output, #range);
#range = (shift #input);
for (#input) {
if ($_ - $range[-1] <= 1) {
push #range, $_ unless $range[-1] == $_; # Prevent repetition
}
else {
push #output, [ #range ];
#range = ($_);
}
}
push #output, [ #range ] if #range;
# Return the result as a string. If a sequence is size 1, then it's just that number.
# Otherwise, it's the first and last number joined by '-'
return join ', ', map { 1 == #$_ ? #$_ : join ' - ', $_->[0], $_->[-1] } #output;
}
print adjacenify( qw/1 2 3 5 7 9 10 11 12 14/ ), "\n";
print adjacenify( 1 .. 5 ), "\n";
print adjacenify( qw/-10 -9 -8 -1 0 1 2 3 5 7 9 10 11 12 14/ ), "\n";
print adjacenify( qw/1 2 4 5 6 7 100 101/), "\n";
print adjacenify( qw/1 62/), "\n";
print adjacenify( qw/1/), "\n";
print adjacenify( qw/1 2/), "\n";
print adjacenify( qw/1 62 63/), "\n";
print adjacenify( qw/-2 0 0 2/), "\n";
print adjacenify( qw/-2 0 0 1/), "\n";
print adjacenify( qw/-2 0 0 1 2/), "\n";
Output:
1 - 3, 5, 7, 9 - 12, 14
1 - 5
-10 - -8, -1 - 3, 5, 7, 9 - 12, 14
1 - 2, 4 - 7, 100 - 101
1, 62
1
1 - 2
1, 62 - 63
-2, 0, 2
-2, 0 - 1
-2, 0 - 2
-2, 0 - 2
And a nice recursive solution:
sub _recursive_adjacenify($$);
sub _recursive_adjacenify($$) {
my ($input, $range) = #_;
return $range if ! #$input;
my $number = shift #$input;
if ($number - $range->[-1] <= 1) {
return _recursive_adjacenify $input, [ #$range, $number ];
}
else {
return $range, _recursive_adjacenify $input, [ $number ];
}
}
sub recursive_adjacenify {
my #input = #_;
# Validate and sort
looks_like_number $_ or
die "Saw '$_' which doesn't look like a number" for #input;
#input = sort { $a <=> $b } #input;
my #output = _recursive_adjacenify \#input, [ shift #input ];
# Return the result as a string. If a sequence is size 1,
# then it's just that number.
# Otherwise, it's the first and last number joined by '-'
return join ', ', map { 2 == #$_ && $_->[0] == $_->[1] ? $_->[0] :
1 == #$_ ? #$_ :
join ' - ', $_->[0], $_->[-1] } #output;
}

Answer 8

Short and sweet Ruby
def range_to_s(range)
return range.first.to_s if range.size == 1
return range.first.to_s + "-" + range.last.to_s
end
def format_ints(ints)
range = []
0.upto(ints.size-1) do |i|
range << ints[i]
unless (range.first..range.last).to_a == range
return range_to_s(range[0,range.length-1]) + "," + format_ints(ints[i,ints.length-1])
end
end
range_to_s(range)
end

Answer 9

My first thought, in Python:
def seq_to_ranges(seq):
first, last = None, None
for x in sorted(seq):
if last != None and last + 1 != x:
yield (first, last)
first = x
if first == None: first = x
last = x
if last != None: yield (first, last)
def seq_to_ranges_str(seq):
return ", ".join("%d-%d" % (first, last) if first != last else str(first) for (first, last) in seq_to_ranges(seq))
Possibly could be cleaner, but it's still waaay easy.
Plain translation to Haskell:
import Data.List
seq_to_ranges :: (Enum a, Ord a) => [a] -> [(a, a)]
seq_to_ranges = merge . foldl accum (id, Nothing) . sort where
accum (k, Nothing) x = (k, Just (x, x))
accum (k, Just (a, b)) x | succ b == x = (k, Just (a, x))
| otherwise = (k . ((a, b):), Just (x, x))
merge (k, m) = k $ maybe [] (:[]) m
seq_to_ranges_str :: (Enum a, Ord a, Show a) => [a] -> String
seq_to_ranges_str = drop 2 . concatMap r2s . seq_to_ranges where
r2s (a, b) | a /= b = ", " ++ show a ++ "-" ++ show b
| otherwise = ", " ++ show a
About the same.

Answer 10

Transcript of an interactive J session (user input is indented 3 spaces, text in ASCII boxes is J output):
g =: 3 : '<#~."1((y~:1+({.,}:)y)#y),.(y~:(}.y,{:y)-1)#y'#/:~"1
g 1 2 3 4 5
+---+
|1 5|
+---+
g 1 2 3 5 7 9 10 11 12 14
+---+-+-+----+--+
|1 3|5|7|9 12|14|
+---+-+-+----+--+
g 12 2 14 9 1 3 10 5 11 7
+---+-+-+----+--+
|1 3|5|7|9 12|14|
+---+-+-+----+--+
g2 =: 4 : '<(>x),'' '',>y'/#:>#:(4 :'<(>x),''-'',>y'/&.>)#((<#":)"0&.>#g)
g2 12 2 14 9 1 3 10 5 11 7
+---------------+
|1-3 5 7 9-12 14|
+---------------+
(;g2) 5 1 20 $ (i.100) /: ? 100 $ 100
+-----------------------------------------------------------+
|20 39 82 33 72 93 15 30 85 24 97 60 87 44 77 29 58 69 78 43|
| |
|67 89 17 63 34 41 53 37 61 18 88 70 91 13 19 65 99 81 3 62|
| |
|31 32 6 11 23 94 16 73 76 7 0 75 98 27 66 28 50 9 22 38|
| |
|25 42 86 5 55 64 79 35 36 14 52 2 57 12 46 80 83 84 90 56|
| |
| 8 96 4 10 49 71 21 54 48 51 26 40 95 1 68 47 59 74 92 45|
+-----------------------------------------------------------+
|15 20 24 29-30 33 39 43-44 58 60 69 72 77-78 82 85 87 93 97|
+-----------------------------------------------------------+
|3 13 17-19 34 37 41 53 61-63 65 67 70 81 88-89 91 99 |
+-----------------------------------------------------------+
|0 6-7 9 11 16 22-23 27-28 31-32 38 50 66 73 75-76 94 98 |
+-----------------------------------------------------------+
|2 5 12 14 25 35-36 42 46 52 55-57 64 79-80 83-84 86 90 |
+-----------------------------------------------------------+
|1 4 8 10 21 26 40 45 47-49 51 54 59 68 71 74 92 95-96 |
+-----------------------------------------------------------+
Readable and elegant are in the eye of the beholder :D
That was a good exercise! It suggests the following segment of Perl:
sub g {
my ($i, #r, #s) = 0, local #_ = sort {$a<=>$b} #_;
$_ && $_[$_-1]+1 == $_[$_] || push(#r, $_[$_]),
$_<$#_ && $_[$_+1]-1 == $_[$_] || push(#s, $_[$_]) for 0..$#_;
join ' ', map {$_ == $s[$i++] ? $_ : "$_-$s[$i-1]"} #r;
}
Addendum
In plain English, this algorithm finds all items where the previous item is not one less, uses them for the lower bounds; finds all items where the next item is not one greater, uses them for the upper bounds; and combines the two lists together item-by-item.
Since J is pretty obscure, here's a short explanation of how that code works:
x /: y sorts the array x on y. ~ can make a dyadic verb into a reflexive monad, so /:~ means "sort an array on itself".
3 : '...' declares a monadic verb (J's way of saying "function taking one argument"). # means function composition, so g =: 3 : '...' # /:~ means "g is set to the function we're defining, but with its argument sorted first". "1 says that we operate on arrays, not tables or anything of higher dimensionality.
Note: y is always the name of the only argument to a monadic verb.
{. takes the first element of an array (head) and }: takes all but the last (curtail). ({.,}:)y effectively duplicates the first element of y and lops off the last element. 1+({.,}:)y adds 1 to it all, and ~: compares two arrays, true wherever they are different and false wherever they are the same, so y~:1+({.,}:)y is an array that is true in all the indices of y where an element is not equal to one more than the element that preceded it. (y~:1+({.,}:)y)#y selects all elements of y where the property stated in the previous sentence is true.
Similarly, }. takes all but the first element of an array (behead) and {: takes the last (tail), so }.y,{:y is all but the first element of y, with the last element duplicated. (}.y,{:y)-1 subtracts 1 to it all, and again ~: compares two arrays item-wise for non-equality while # picks.
,. zips the two arrays together, into an array of two-element arrays. ~. nubs a list (eliminates duplicates), and is given the "1 rank, so it operates on the inner two-element arrays rather than the top-level array. This is # composed with <, which puts each subarray into a box (otherwise J will extend each subarray again to form a 2D table).
g2 is a mess of boxing and unboxing (otherwise J will pad strings to equal length), and is pretty uninteresting.

Answer 11

Here's my Haskell entry:
runs lst = map showRun $ runs' lst
runs' l = reverse $ map reverse $ foldl newOrGlue [[]] l
showRun [s] = show s
showRun lst = show (head lst) ++ "-" ++ (show $ last lst)
newOrGlue [[]] e = [[e]]
newOrGlue (curr:other) e | e == (1 + (head curr)) = ((e:curr):other)
newOrGlue (curr:other) e | otherwise = [e]:(curr:other)
and a sample run:
T> runs [1,2,3,5,7,9,10,11,12,14]
["1-3","5","7","9-12","14"]

Answer 12

Erlang , perform also sort and unique on input and can generate programmatically reusable pair and also a string representation.
group(List) ->
[First|_] = USList = lists:usort(List),
getnext(USList, First, 0).
getnext([Head|Tail] = List, First, N) when First+N == Head ->
getnext(Tail, First, N+1);
getnext([Head|Tail] = List, First, N) ->
[ {First, First+N-1} | getnext(List, Head, 0) ];
getnext([], First, N) -> [{First, First+N-1}].
%%%%%% pretty printer
group_to_string({X,X}) -> integer_to_list(X);
group_to_string({X,Y}) -> integer_to_list(X) ++ "-" ++ integer_to_list(Y);
group_to_string(List) -> [group_to_string(X) || X <- group(List)].
Test getting programmatically reusable pairs:
shell> testing:group([34,3415,56,58,57,11,12,13,1,2,3,3,4,5]).
result> [{1,5},{11,13},{34,34},{56,58},{3415,3415}]
Test getting "pretty" string:
shell> testing:group_to_string([34,3415,56,58,57,11,12,13,1,2,3,3,4,5]).
result> ["1-5","11-13","34","56-58","3415"]
hope it helps
bye

Answer 13

VBA
Public Function convertListToRange(lst As String) As String
Dim splLst() As String
splLst = Split(lst, ",")
Dim x As Long
For x = 0 To UBound(splLst)
Dim groupStart As Integer
groupStart = splLst(x)
Dim groupEnd As Integer
groupEnd = groupStart
Do While (x <= UBound(splLst) - 1)
If splLst(x) - splLst(x + 1) <> -1 Then Exit Do
groupEnd = splLst(x + 1)
x = x + 1
Loop
convertListToRange = convertListToRange & IIf(groupStart = groupEnd, groupStart & ",", groupStart & "-" & groupEnd & ",")
Next x
convertListToRange = Left(convertListToRange, Len(convertListToRange) - 1)
End Function
convertListToRange("1,2,3,7,8,9,11,12,99,100,101")
Return: "1-3,7-9,11-12,99-101"