The first, the last, and everything between

Question

Given two integers, output the two integers, and then the range between them (excluding both).

The order of the range must be the same as the input.

Examples:

 Input        Output
 0,  5   ->   [0, 5, 1, 2, 3, 4]
-3,  8   ->   [-3, 8, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7]
 4,  4   ->   [4, 4]
 4,  5   ->   [4, 5]
 8,  2   ->   [8, 2, 7, 6, 5, 4, 3]
-2, -7   ->   [-2, -7, -3, -4, -5, -6]

Answer 1

QBASIC, 39 53 bytes

INPUT a,b
?a
?b
FOR q=a+1TO b-1 STEP SGN(b-a)
?q
NEXT

Added the STEP parameter to account for a>b, and that uses the SGN() function to get a -1 or a +1 as increment. This however breaks the REPL because the SGN() function isn't implemented there...

Try it (the old answer) online!

Answer 2

C (gcc), 65 bytes

f(a,b){for(printf("%d %d",a,b);a<b?++a<b:--a>b;)printf(" %d",a);}

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Not very exciting. The loop increment is borrowed from an early version of Kevin Cruijssen's Java answer.

Answer 3

Powershell, 41 bytes

param($a,$b)$a;$b;$a..$b|?{$_-notin$a,$b}

Less golfed test script:

$f = {

param($a,$b)
$a                # push $a to a pipe
$b                # push $b to a pipe
$a..$b|?{         # push to pipe all integers from $a to $b
    $_-notin$a,$b # ...except $a and $b itself
}

}

@(
    ,( 0,  5  ,   0, 5, 1, 2, 3, 4)
    ,(-3,  8  ,   -3, 8, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7)
    ,( 4,  4  ,   4, 4)
    ,( 4,  5  ,   4, 5)
    ,( 8,  2  ,   8, 2, 7, 6, 5, 4, 3)
    ,(-2, -7  ,   -2, -7, -3, -4, -5, -6)
) | % {
    $a,$b,$expected = $_
    $result = &$f $a $b
    "$("$result"-eq"$expected"): $result"
}

Output:

True: 0 5 1 2 3 4
True: -3 8 -2 -1 0 1 2 3 4 5 6 7
True: 4 4
True: 4 5
True: 8 2 7 6 5 4 3
True: -2 -7 -3 -4 -5 -6

Explanation:

The basic concept of Powershell is the pipe. Pipe is an array. All results that push into the pipe fall into the array. So we should just push the values into the pipe in the correct order.

Answer 4

Java, 739, 555 bytes

 public static List<Integer> firstLastAndEverythingBetween(final int a, final int b) { if (a == b) { return addAB(a, b); } final List<Integer> result = addAB(a, b); int initial = getInitial(a, b); for (int n = 1; n < Math.abs(b - a); n++) { result.add(initial); if (b > a) { initial++; } else { initial--; } } return result; } private static int getInitial(int a, int b) { return (b > a) ? (a + 1) : (a - 1); } private static List<Integer> addAB(int a, int b) { final List<Integer> result = new ArrayList<>(); result.add(a); result.add(b); return result; }

Answer 5

Python 3, 53 48 bytes

-5 bytes thanks to @mypetlion.

lambda a,b:[a,b,*range(a+1-2*(a>b),b,1-2*(a>b))]

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Answer 6

APL(NARS), 14 chars, 28 bytes

{⍺⍵,¯1↓1↓⍺..⍵}

test

  f←{⍺⍵,¯1↓1↓⍺..⍵}
  ¯3 f 8
¯3 8 ¯2 ¯1 0 1 2 3 4 5 6 7 
  0 f 5      
0 5 1 2 3 4 

Answer 7

C (clang), 60 bytes

f(a,b,c)int*c;{for(*c++=a,*c++=b;(a<b?++a<b:--a>b);*c++=a);}

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Answer 8

C++, 143 140 bytes

-3 bytes thanks to Zacharý

#define V std::vector<int
#include<vector>
V>f(int a,int b){V>v{a,b};int c=2*(b>a)-1;for(int i=a+c;a!=b&&b!=i;i+=c)v.push_back(i);return v;}

How to use :

In the main function :

std::cout << f(0, 5) << '\n';
std::cout << f(-3, 8) << '\n';
std::cout << f(4, 4) << '\n';
std::cout << f(4, 5) << '\n';
std::cout << f(8, 2) << '\n';
std::cout << f(-2, -7) << '\n';

also add :

template<typename T>
std::ostream& operator<<(std::ostream& os, const std::vector<T>& v) {
    os << "[ ";
    for (const auto& a : v) {
        os << a << ' ';
    }
    return os << ']';
}

Answer 9

Haskell, 18 34 bytes as function, 46 52 as an IO action, 56 62 as program

x!y=x:y:[x+1..y-1]++[x-1,x-2..y+1]

or, as an IO action to take input, there are these options (all the same length):

main=(\x y->x:y:[x+1..y-1]++[x-1,x-2..y+1])<$>readLn<*>readLn
--
x!y=x:y:[x+1..y-1]++[x-1,x-2..y+1]
main=(!)<$>readLn<*>readLn
--
f x y=x:y:[x+1..y-1]++[x-1,x-2..y+1]
main=f<$>readLn<*>readLn

or, as an IO action taking input and printing the result:

main=((\x y->x:y:[x+1..y-1]++[x-1,x-2..y+1])<$>readLn<*>readLn)>>=print
--
x!y=x:y:[x+1..y-1]++[x- 1,x-2..y+1]
main=((!)<$>readLn<*>readLn)>>=print
--
f x y=x:y:[x+1..y-1]++[x-1,x-2..y+1]
main=(f<$>readLn<*>readLn)>>=print

The last one is also available on Try It Online! (You can try it online there)
Please remember that TIO requires the input be entered beforehand in the "Input" box.

De-golfed:

enumerate' x y = x : y : [x + 1 .. x - 1] ++ [x - 1 , x - 2 .. y + 1]
main = do x <- readLn
          y <- readLn
          print (enumerate' x y)

Note: : is an operator to prefix an element to a list.

Note: a!b will not work if -XBangBatterns is enabled.
Use a#b instead.

Note: -XNegativeLiterals will break this. Insert a space between x- and 1.
This will parse as x + (negate 1) instead of x $ (-1).

Answer 10

C# (Visual C# Interactive Compiler), 86 bytes

x=>y=>{var l=new List<int>{x,y};for(int s=x>y?-1:1;x!=y&(x+=s)!=y;)l.Add(x);return l;}

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-1 thanks @auhmaan!

Answer 11

JavaScript, 75 bytes

(a,b,n=-~(a>b?a-b:b-a)-2)=>[a,b,...[...Array(n<2?0:n)].map(_=>a<b?++a:--a)]

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Answer 12

Husk, 5 bytes

+¹ht…

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Thanks to Dominic Van Essen for the fix.

Answer 13

Tcl, 85 bytes

proc R a\ b {puts $a\n$b
if $a>$b {lassign $a\ $b b a}
while {[incr a]<$b} {puts $a}}

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Answer 14

JavaScript (ES6), 97 bytes

a=>b=>[a,b,...((c,d)=>c==d?[]:(s=(m=Math).min(a,b),Array(m.abs(a-b)-1).fill().map(n=>++s)))(a,b)]

Can probably be improved!!

Answer 15

Python 3, 57 bytes

def m(a,b):s=(1,-1)[a>b];return[a,b]+list(range(a+s,b,s))

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Answer 16

JAVA 147 bytes

f(int a, int b){System.out.print(a+""+b);if(a<b){for(int i=a+1; i<b;i++){System.out.print(i);}}else{for(int i=a-1;i>b;i--){System.out.print(i);}}}

Answer 17

Racket, 68 bytes

(define(f a b)(let([s(if(> a b)-1 1)])(list* a b(range(+ a s)b s))))

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The same as TheGreatGeek's Clojure solution

Answer 18

perl -M5.010 -alE, 58 bytes

@_=($F[0]..$F[1],reverse$F[1]..$F[0]);say"$_ @_[1..$#_-1]"

This reads two integers (on a single line, separated by whitespace) from STDIN and writes the result to STDOUT.

Answer 19

Bash, 62 bytes

a=($(seq $1 $2|tr $'\n' ' '))
echo $@ ${a[@]:1:((${#a[@]}-2))}

explanation

seq does we want but produces lines, tr turns the lines into space-delimited output, the outer parens make that into an array. $@ is the original input, and the rest of it slices the 0-indexed array from 1 and takes the next "array len - 2" chars.

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(NOTE: not all test cases work on TIO for reasons i don't understand, but they do work on my local machine)

Answer 20

C# (.NET Core), 99 bytes

a=>b=>new[]{a,b}.Concat(b>a?Enumerable.Range(a+1,b+~a):Enumerable.Range(b+1,a!=b?a+~b:0).Reverse())

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First (and probably last) time golfing in C#.

Answer 21

F# (Mono), 68 bytes

let f x y=
 let s=if x<y then 1 else-1
 Seq.append[x;y][x+s..s..y-s]

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Answer 22

C# .NET Core, 56 441 bytes

Here is my corrected solution without predefined input variables and without string joining.

(a,b)=>n(a,b).c(k(a-b)>1?b>a?r(a+1,b-a-1):r(b+1,a-b-1).i():n());

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n(a,b)               => returns new array with a and b in it
  c(                 => concat the following content
    k(a-b)>1         => when absolute value of a-b greater then one
      b>a            => when b bigger a
        r(a,b-a)     => range from a to b-a
        r(b,a-b).i() => else range from b to a-b reversed
      n()            => else empty array

I've created and used the following declarations. When I include those and the minimum code required to compile in the byte count I get 640 bytes.

public static class Extensions{
  public static IEnumerable<T> c<T>(this IEnumerable<T> a, IEnumerable<T> b) => a.Concat(b);
  public static IEnumerable<T> i<T>(this IEnumerable<T> a) => a.Reverse();
}

public static Func<int, int> k = Math.Abs;
public static Func<int, int, IEnumerable<int>> r = Enumerable.Range;
public static int[] n(params int[] a) => a;

---

using System.Linq;using System.Collections.Generic;using System;public static class E{public static IEnumerable<T> c<T>(this IEnumerable<T> a,IEnumerable<T> b)=>a.Concat(b);public static IEnumerable<T> i<T>(this IEnumerable<T> a)=>a.Reverse();}public class P{public static Func<int,int> k=Math.Abs;public static Func<int,int,IEnumerable<int>> r=Enumerable.Range;public static int[] n(params int[] a)=>a;public static void Main(string[] args){Console.WriteLine($"[ {string.Join(", ", f(int.Parse(args[0]), int.Parse(args[1])))} ]");}public static Func<int,int,IEnumerable<int>> f=(a,b)=>n(a,b).c(k(a-b)>1?b>a?r(a,b-a):r(b,a-b).i():n());}

Answer 23

Vyxal, 9 bytes

=[‿|₌‿⁂JU

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Explained

The output for this challenge is the two input numbers paired together joined with the range between those two numbers. However, there can't be any duplicates in the joined list, so it needs to be uniquified by order of appearance. But we also need to make sure that the two numbers aren't the same: otherwise, a singleton will be returned.

Answer 24

F#, 67 bytes

let f x y=if(y-x<0)then[x;y]@List.rev[y+1..x-1]else[x;y]@[x+1..y-1]

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Appends a list containing x and y and a list iterating from x+1 to y-1 (if positive) or a list in reverse iteration order from y+1 to x-1 (if negative)

Answer 25

JavaScript (ES6), 76 bytes

(a,b,c=[a,b])=>{c[c.length]=a<b?++a:--a;return a==b?(c.pop()+1&&c):f(a,b,c)}

invocation should not pass the third argument: f(-9,9)

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im still thinking...

Answer 26

Perl 5 -pl, 46 bytes

/ /;$_.=join$",'',$`+1..$'-1,reverse$'+1..$`-1

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