Adjacent number algorithm grouper

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

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 2

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 3

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 4

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 5

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 6

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 7

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 8

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 9

Thanks for a lot of great thoughts!

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

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 13

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