Union of Intervals

Question

Given a list of intervals, perform the union of them, and reduce overlaps. That means, overlapping parts are reduced. ([a, b] U [c, d] = [a, d] if b > c) Assuming all a < b in all intervals [a, b]. Implement as a function of a list of input intervals -> list of output intervals. Shortest code wins. You cannot use any existing libraries.

Clarifications:

• Open and closed intervals are not distinguished.
• Intervals for real numbers, not integers. ([2, 3], [4, 5] -> [2, 3], [4, 5])
• No need to sort output intervals
• The order if inputs do not matter
• Illegal inputs are only [a, b] where b >= a, it has nothing to do with the order of input intervals and the number of input intervals.
• You do not need to show an error message on undefined behaviors

Examples (with number lines)

 [2, 4], [7, 9] --> [2, 4], [7, 9]
   234
        789
-> 234  789

 [1, 5], [2, 10] --> [1, 10] (overlapping [2, 5] reduced)

   12345
    234567890
-> 1234567890
 [2, 4], [3, 6], [8, 9] -> [2, 6], [8, 9]
   234
    3456
         89
-> 23456 89

 [4, 2], [2, 2] -> (undefined behavior: against the assumption)

Answer 1

Haskell (103)

I think, it is way too verbose for Haskell. Thanks to Hoa Long Tam for his sorting function.

m%(x:y)|x>m=m:x:y|2>1=x:m%y;m%_=[m]
(x:y)?l|x`elem`l=y?l|0<1=x:y?(x:l);a?_=a
a∪b=foldr(%)[](a++b)?[]

In Haskell, an intervall from a to b is denoted by [a..b]. My notation is very similar to the mathematical notation. Use it like this:

[a..b] ∪ [c..d] ∪ ... ∪ [y..z]

Answer 2

Ok, here's my 250 character crack at it.

void n(int a[]){if(!a[2])return;if(a[2]<=a[1]){if(a[1]<a[3])a[1]=a[3];
int *b=a+2;while(*b=*(b+2))++b;n(a);}n(a+2);}
void m(int a[]){if(!a[2])return;if(a[0]>a[2]){int s=a[0],t=a[1];
a[0]=a[2];a[2]=s;a[1]=a[3];a[3]=t;m(a+2);m(a);n(a);}m(a+2);n(a+2);}

The function takes an int array, and operates on it in-situ. The array is terminated by a 0, and the intervals may be given in any order.

Sample output:

input list: (7,9) (5,6) (1,4) (15,18) (13,16) (2,3) (8,11) 
output list: (1,4) (5,6) (7,11) (13,18) 

Sample program:

#include <stdio.h>

void n(int a[]){if(!a[2])return;if(a[2]<=a[1]){if(a[1]<a[3])a[1]=a[3];
int *b=a+2;while(*b=*(b+2))++b;n(a);}n(a+2);}
void m(int a[]){if(!a[2])return;if(a[0]>a[2]){int s=a[0],t=a[1];
a[0]=a[2];a[2]=s;a[1]=a[3];a[3]=t;m(a+2);m(a);n(a);}m(a+2);n(a+2);}


/*
void n(int a[])
{
    if(!a[2])return;
    if(a[2]<=a[1]) {
        if(a[1]<a[3])
            a[1]=a[3];
        int *b=a+2;
        while(*b=*(b+2))++b;
        n(a);
    }
    n(a+2);
}

void m(int a[])
{
    if(!a[2])return;
    if(a[0]>a[2]) {
        int s=a[0],t=a[1];
        a[0]=a[2];a[2]=s;
        a[1]=a[3];a[3]=t;
        m(a+2);m(a);n(a);
    }
    m(a+2);n(a+2);
}
*/

void p(int a[]) 
{
    if(!*a) {
        printf("\n");
        return;
    }
    printf("(%d,%d) ",a[0],a[1]);
    p(a+2);
}

int main (int argc, const char * argv[]) 
{
    // Code golf entry
    // Interval Merging

    int a[] = {7,9,5,6,1,4,15,18,13,16,2,3,8,11,0};
    printf( "input list: " ); p(a);
    m(a);
    printf( "output list: " ); p(a);

    return 0;
}

Answer 3

GolfScript, 32

[{1$1$*-2%~2->{*$[(\)\;]}{}if}*]

• Add 2 characters if you prefer a block, 4 if you prefer a named block.
• Input and output are array of pairs, e.g. [[2 4] [3 5]]
• Assumes that input is ordered by the first element.
• Compacts "adjacent" ranges ([2 4][5 6] -> [2 6])
• First GolfScript effort. Advice & rotten fruit appreciated.

Full test program:

[~](;2/[{1$1$*-2%~2->{*$[(\)\;]}{}if}*]`

Example invocation:

ruby golfscript.rb intervals.gs <<EOF
3
2 4
3 6
8 9
EOF
# Expected output: [[2 6] [8 9]]

Answer 4

Python, 100 chars

def f(L):R=sorted(set(p for p in sum(L,[])if 1-any(x<p<y for x,y in L)));return zip(R[::2],R[1::2])
print f([[2, 4], [7, 9]])
print f([[1, 5], [2, 10]])
print f([[3, 6], [2, 4], [8, 9]])
print f([[1, 5], [3, 5], [4, 5]])

generates

[(2, 4), (7, 9)]
[(1, 10)]
[(2, 6), (8, 9)]
[(1, 5)]

Takes all the endpoints of the intervals, removes any that are strictly inside another interval, uniquifies & sorts them, and pairs them up.

Answer 5

Haskell, 55 characters

v(q@(a,b):p@(c,d):r)|c>b=q:v(p:r)|1<3=v((a,d):r);v x=x

If the input is unsorted, then 88 characters:

p@(a,b)§(q@(c,d):r)|b<c=p:q§r|a>d=q:p§r|1<3=(min a c,max b d)§r;p§_=[p]
u i=foldr(§)[]i

Test runs:

ghci> testAll v
pass: [(2,4),(7,9)] --> [(2,4),(7,9)]
pass: [(1,5),(2,10)] --> [(1,10)]
pass: [(2,4),(3,6),(8,9)] --> [(2,6),(8,9)]
ghci> testAll u
pass: [(2,4),(7,9)] --> [(2,4),(7,9)]
pass: [(1,5),(2,10)] --> [(1,10)]
pass: [(2,4),(3,6),(8,9)] --> [(2,6),(8,9)]

I'm assuming that "can't use any existing libraries" precludes importing List and calling sort. If that were legal, than the unsorted version would be only 71 characters.

Answer 6

Scala, 272 characters

type p=List[(Int,Int)];def f(l:p):p={var(a,s,c,o)=(new Array[Int]((l map(x=>x._2)max)+1),0,0,List[Int]());l map(x=>(a(x._1)+=1,a(x._2)-=1));while(c<a.size){s+=a(c);if(a(c)==1&&s==1)o=o:+c;if(a(c)== -1&&s==0)o=o:+c;c+=1};return(o.grouped(2).map(x=>(x.head,x.last)).toList)}

Usage:

object Intervals2 extends Application
{
    type p=List[(Int,Int)];def f(l:p):p={var(a,s,c,o)=(new Array[Int]((l map(x=>x._2)max)+1),0,0,List[Int]());l map(x=>(a(x._1)+=1,a(x._2)-=1));while(c<a.size){s+=a(c);if(a(c)==1&&s==1)o=o:+c;if(a(c)== -1&&s==0)o=o:+c;c+=1};return(o.grouped(2).map(x=>(x.head,x.last)).toList)}

    print(f(List((1,2),(3,7),(4,10))))
}

Creates an array and inserts a 1 for every interval start and a -1 for every interval end. Then steps through the array adding the values to a counter outputting a start every time the counter steps from 0 to 1 and an end when it steps from 1 to 0. Probably unnecessarily complicated.

Output:

List((1,2), (3,10))

Answer 7

Perl (146) (92) (90)

golfed down to 90 chars, creatively using the regex engine

sub u{map$h[$_]=1,@$_[0]..@$_[1]for@_;$w.=$_+0for@h;push@r,$-[0],$+[0]-1while$w=~/1+/g;@r}

usage example:

my @out1 = u([1, 5], [2, 10]); # (1,10)
my @out2 = u([2, 4], [3, 6], [8, 9]); # (2, 6, 8, 9)

let's explain this code a bit.

this subroutine receives an array of arrayrefs, each aref pointing to an array containing two elements, start and end of the interval: ([2, 4], [3, 6], [8, 9])

for every aref, we generate an array of elements from first to last ($_->[0] .. $_->[1]). then we use map to set elements of such indexes in @h to 1.

for (@_) {
    map {$h[$_] = 1} ($_->[0] .. $_->[1]);
}

after this, @h will contain either ones (for intervals) or undefs, depicted below as hyphens for clarity.

index: 0 1 2 3 4 5 6 7 8 9
@h:    - - 1 1 1 1 1 - 1 1

next we build a string from @h, adding 0 to replace undefs with something more useful (undef + 0 = 0).

$w .= $_+0 for @h;

$w contains 011111011 now.

it's time to abuse the regex engine a bit.

push @r, ($-[0], $+[0]-1) while $w=~/1+/g;

after successful matches, @- and @+ arrays contain respectively position of start and end of each match; 0th element is used for the entire match, first for $1, second for $2 and so on.

$+[0] actually contains the position of first non-matching char, so we have to substract one.

@r contains (2, 6, 8, 9) now.

@r

to make the sub return @r.

Answer 8

Scala 305 279 chars without invocation:

type I=(Int,Int)
def l(p:I,q:I)=if(p._1<q._1)true else if(p._1>q._1)false else p._2<q._2
def r(l:List[I]):List[I]=l match{case x::y::z=>{if(y._1<=x._2&&y._2>x._2)(x._1,y._2)::r(z)else
if(y._1<=x._2&&y._2<=x._2)x::r(z)else  
x::r(y::z)}case _=>l}
def c(v:List[I])=r(v.sortWith(l))

invocation:

val i=List((7,9),(5,6),(1,4),(15,18),(13,16),(2,3),(8,11))
c(i)
res0: List[(Int, Int)] = List((1,4), (5,6), (7,11), (13,18))

Answer 9

Brachylog, 12 bytes

⟦₂ᵐcod~c~⟦₂ᵐ

Try it online!

A delightfully declarative solution, taking input as a list of lists through the input variable and outputting a list of lists through the output variable.

        ~⟦₂ᵐ    The output is a list of intervals, where each interval is a range in
      ~c        the smallest partition of
  ᵐ             each element of the input
⟦₂              converted to an inclusive range,
   c            concatenated,
    o           sorted,
     d          and deduplicated
        ~⟦₂ᵐ    for which each element of the partition is a range.

Answer 10

Clojure, 138 bytes

#(let[S(set(apply mapcat range(apply map list %)))Q(sort S)](map list(for[s Q :when(not(S(dec s)))]s)(for[s(map inc Q):when(not(S s))]s)))

This shortens to 119 bytes if the input is more flexible, namely a list of intervals' starting points AND a list of intervals' ending points:

#(let[S(set(mapcat range % %2))Q(sort S)](map list(for[s Q :when(not(S(dec s)))]s)(for[s(map inc Q):when(not(S s))]s)))

There has to be a better way.

Answer 11

Japt, 18 bytes

This feels far too long. I/O as sorted, 2D array of intervals.

c_ròÃòÈaY Éîg[TJ]

Try it

Answer 12

Jelly (fork), 11 bytes

r/€FṢQŒġ.ịⱮ

Try it online! or rather, don't

Outputs the ranges as [b, a] where \$b > a\$ (so the example [2, 4], [3, 6], [8, 9] is output as [6, 2], [9, 8])

How it works

r/€FṢQŒġ.ịⱮ - Main link. Takes a list of pairs on the left
  €         - Over each pair:
r/          -   Reduce into a range
   FṢQ      - Flatten, sort and deduplicate
      Œġ    - Group adjacent consecutive elements
        .ịⱮ - Select the first and last from each group

Answer 13

Perl 5 -MList::Util=max -n, 89 bytes

@r=/\S+/g;map{/ /;$`<=$r[1]?$r[1]=max@r,$'*1:(say"@r")|(@r=($`,$'))}sort{$a-$b}<>;say"@r"

Try it online!

Answer 14

Ruby, 75 bytes

->a{a.map{[*_1.._2]}.reduce(:|).sort.slice_when{_2-_1>1}.map{|a,*,b|[a,b]}}

Attempt This Online!

Answer 15

Mathematica, 18 bytes

List@@Interval@@#&

View it on Wolfram Cloud!