Find the temperature closest to 0

Question

In this exercise, you have to analyze records of temperature to find the closest to zero.

Write a program that prints the temperature closest to 0 among input data.

Input

• N, the number of temperatures to analyse (optional). This will be nonzero.
• The N temperatures expressed as integers ranging from -273 to 5526.

Output

Output the temperature closest to 0. If two temperatures are equally close, take the positive one. For instance, if the temperatures are -5 and 5, output 5.

Example

Input

    5
    1 -2 -8 4 5

Output

    1

This challenge is similar to this one on CodinGame, you can view the problem statement source here. Some modifications have been made to the text.

Answer 1

Python, 35 bytes

lambda l:max((-x*x,x)for x in l)[1]

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Narrowly beats:

lambda l:min(l,key=lambda x:2*x*x-x)
lambda l:min(sorted(l)[::-1],key=abs)

Answer 2

JavaScript (ES6), 33 bytes

a=>a.reduce((m,n)=>n*n-n<m*m?n:m)

Demo

let f =

a=>a.reduce((m,n)=>n*n-n<m*m?n:m)

console.log(f([1, -2, -8, 4, 5]))

Answer 3

Jelly, 4 bytes

AÐṂṀ

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How it works

AÐṂṀ  Main link. Argument: A (array)

AÐṂ   Take all elements with minimal absolute value.
   Ṁ  Take the maximum. Yields 0 for an empty list.

Answer 4

Brachylog (2), 5 or 2 bytes

~{≜∈}

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It wouldn't surprise me if there were a shorter solution along the lines of the Jelly or Pyke, but I like this one because it's so weird.

This is simply Brachylog's "find a list containing the input" operator ∈, with an evaluation strategy ≜ that simply tries explicit values for integers until it finds one that works (and it happens to try in the order 0, 1, -1, 2, -2, 3, etc., which is surprisingly handy for this problem!), and inverted ~ (so that instead of trying to find an output list containing the input, it's trying to find an output contained in the input list). Unfortunately, inverting an evaluation strategy along with the value itself costs 2 bytes, so this doesn't beat the Jelly or Pyke solutions.

There's also a dubious 2-byte solution ≜∈. All Brachylog predicates take exactly two arguments, which can each be inputs or outputs; by convention, the first is used as an input and the second is used as an output, but nothing actually enforces this, as the caller has complete control over the argument pattern used (and in Prolog, which Brachylog compiles to, there's no real convention about argument order). If a solution which takes input through its second argument and produces output through its first argument is acceptable, there's no requirement to do the inversion.

Answer 5

R, 31 bytes

(l=scan())[order(abs(l),-l)][1]

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Answer 6

Pyke, 4 bytes

S_0^

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S_   - reversed(sorted(input))
  0^ - closest_to(^, 0)

Answer 7

Mathematica, 21 19 bytes

Max@*MinimalBy[Abs]

Composition of functions. Takes a list of integers as input and returns an integer as output.

Answer 8

MATL, 9 6 bytes

0iSPYk

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Thanks to Luis Mendo for pointing out there's actually a built-in for this in MATL, which is not present in MATLAB.

 i     % Take input
  S    % Sort
   P   % Reverse to have largest value first.
0   Yk % Closest value to zero, prioritizing the first match found

If you wish to follow the spec and also provide the number of temperatures to be read, this should be given as the second input, and will be silently discarded.

Answer 9

C++14, 64 bytes

As unnamed lambda, expecting first argument to be like vector<int> and returing via reference parameter.

[](auto L,int&r){r=L[0];for(int x:L)r=x*x<=r*r?r+x?x:x>r?x:r:r;}

Ungolfed and usage:

#include<iostream>
#include<vector>

auto f=
[](auto L,int&r){ 
  r=L[0];
  for(int x:L)
    r = x*x<=r*r ?   //x is absolute less or equal
      r+x ?          //r==-x?
        x :          //no, just take x
        x>r ?        //take the positive one
          x :
          r      :
      r              //x was absolute greater, so keep r
    ;
}
;

int main(){
 std::vector<int> v = {5, 2, -2, -8, 4, -1, 1};
 int r;
 f(v,r);
 std::cout << r << std::endl;
}

Answer 10

Perl 5 -p, 21 bytes

Added +1 for -p since this challenge precedes "options don't count"

$G[abs]=$_}{$_+="@G"

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Answer 11

Japt -g, 3 bytes

ña¼

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Japt, 5 bytes

ña¼ v

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Answer 12

05AB1E, 16 bytes

Îåà_Di¹{RDÄßs\kè

Explanation:

Î                 Push 0 and [implicit] input array
 å                For each element in the array, is it a 0?
  à               Find the greatest value (1 if there was a 0 in it, 0 otherwise)
   _              Negate boolean (0 -> 1, 1 -> 0)
    D             Duplicate
     i            If true, continue
      ¹           Push input
       R          Reversed
        D         Duplicated
         Ä        Absolute value
          ß       Greatest element
           s      Swap top two values in stack
            \     Delete topmost value in stack
             k    Index of greatest element in the array
              è   Value of the index in the input array

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Not at short as I'd like, but at least it works.

Answer 13

Haskell, 29 bytes

snd.maximum.map(\x->(-x^2,x))

Usage example: snd.maximum.map(\x->(-x^2,x)) $ [1,-2,-8,4,5]-> 1.

Map each element x to a pair (-x^2,x). Find the maximum and pick the 2nd element from the pair.

Answer 14

PHP, 81 bytes

two solutions: the first one requires an upper bound; both require all temperatures nonzero.

for($r=9e9;$t=$argv[++$i];)abs($t)>abs($r)|abs($t)==abs($r)&&$t<$r?:$r=$t;echo$r;
for(;$t=$argv[++$i];)$r&&(abs($t)>abs($r)|abs($t)==abs($r)&&$t<$r)?:$r=$t;echo$r;

Run with -r, provide temperatures as command line arguments.

A lazy solution for 92 bytes:

$a=array_slice($argv,1);usort($a,function($a,$b){return abs($a)-abs($b)?:$b-$a;});echo$a[0];

defines a callback function for usort and uses it on a slice of $argv.

Answer 15

Perl 6, 15 bytes

*.min:{.²,-$_}

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Answer 16

Java 8, 63 39 bytes

s->s.reduce(1<<15,(r,i)->r*r-r<i*i?r:i)

Port from @Arnauld's JavaScript answer. I couldn't find anything shorter..

-10 bytes converting Java 7 to 8,
and another -24 bytes thanks to @OlivierGrégoire.

Explanation:

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s->                   // Method with IntStream parameter and int return-type
  s.reduce(1<<15,     //  Start `r` at 32768 (`32768^2` still fits inside a 32-bit integer)
     (r,i)->r*r-r<i*i?//  If `r^2 - r` is smaller than `i^2`:
       r              //   Leave `r` the same
      :               //  Else:
       i)             //   Replace the current `r` with `i`
                      //  Implicitly return the resulting `r`

Answer 17

JavaScript (Node.js), 30 bytes

a=>a.sort((a,b)=>a*a-a>b*b)[0]

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Answer 18

Octave, 28 bytes

@(x)max(x(min(a=abs(x))==a))

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Finds the smallest absolute value (min(a=abs(x))), then maps it back to which elements have that smallest absolute value in case of a tie (x(...==a)). Finally takes the maximum value of the results (max(...)) to ensure that in the event of a tie we get one result which is the positive one.

Answer 19

PHP, 209 Bytes

Not really golfed, but wanted to try to use mostly array functions, avoiding loops

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Code

function f($a){$a[]=0;sort($a);$k=array_search(0,$a);$o=array_slice($a,$k+1);$u=array_reverse(array_diff($a,$o));if($u[1]&&$o[0]){$r=(abs($u[1])<abs($o[0]))?$u[1]:$o[0];}else{$r=($u[1])?$u[1]:$o[0];}return$r;}

Explanation

function f($a){
$a[]=0;                                #As all nonzero values, 0 is pushed
sort($a);                              #Sorting the array
$k=array_search(0,$a);                 #Search where zero is
$o=array_slice($a,$k+1);               #Array with all values >0
$u=array_reverse(array_diff($a,$o));   #Array with all smaller than zero values
                                       #Array has ascending order so it gets reversed              
if($u[1]&&$o[0]){                      #if both are set we just compare the abs value
    $r=(abs($u[1])<abs($o[0]))?$u[1]:$o[0];
}else{
    $r=($u[1])?$u[1]:$o[0];            #al least one of them will allways be set
                                       #the first value is always returned
                                       #the 0 value on $u is not usetted that why $u[1]
}
return $r;
}

Answer 20

Julia 0.6, 29 bytes

x->sort(x,by=x->abs(x-.1))[1]

Sort by the absolute value of the temperature, but subtract .1 first to ensure positive temperatures always win.

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Answer 21

05AB1E, 5 bytes

{R0.x

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Or alternatively:

ÄWQÏà

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Explanation:

{      # Sort the (implicit) input-list lowest-to-highest
 R     # Reverse it to highest-to-lowest
  0.x  # Pop and push the item closest to 0
       # (after which the result is output implicitly)

Ä      # Get the absolute value of each in the (implicit) input-list
 W     # Push the minimum (without popping the list)
  Q    # Check which are equal to this minimum
   Ï   # Only keep the values in the (implicit) input-list at the truthy indices
    à  # Pop and push the maximum of the remaining values
       # (after which the result is output implicitly)

Answer 22

Thunno 2 h, 4 bytes

ṠrþA

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Explanation

ṠrþA  # Implicit input
Ṡr    # Reverse-sort
  þA  # Sort by |x|
      # Take the first item
      # Implicit output

Answer 23

Vyxal G, 20 bits^v2, 2.5 bytes

⁽ȧP

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Finds the minimum in the list by absolute value

Answer 24

Python, 62 bytes

def f(n):
    r=min(map(abs,n))
    return r if r in n else -r

It feels like there is a much simpler/shorter solution.

Answer 25

C (clang), 74 bytes

k,m;f(*i,n){for(k=*i;--n>0;k=(k*k>m*m||(k==-m&&k<m))?m:k)m=i[n];return k;}

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Answer 26

Stax, 6 bytes

¢M║⌠§☺

Run and debug it at staxlang.xyz!

Unpacked (6 bytes) and explanation

{|a}eoH
{  }e      Get all elements of array producing the minimum value from a block.
 |a          Absolute value.
     o     Sort ascending.
      H    Take last. Implicit print.

Answer 27

Pyth, 4 bytes

.m.a

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Explanation:

.m.a   #Code
.m.abQ #With implicit variables
.m   Q #Filter input list for values b with minimal value of the following:
  .ab  #Absolute value of b

Answer 28

Python 3, 31 bytes

a=lambda q:sorted(q,key=abs)[0]

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Sorts items by absolute value and prints the first element.

Answer 29

Avail, 88 bytes

Method"f_«_»"is[c:[1..∞),t:integer*|quicksort a,b in t[1..c] by[|a-0.1|<|b-0.1|][1]]

Approximate deobfuscation:

Method "nearest to zero among first _ of «_»" is
[
    count : natural number,
    temps : integer*
|
    (quicksort a, b in temps[1..count] by [|a-0.1| < |b-0.1|])[1]
]

The pattern syntax in the method's name, f_«_», will read the initial count as a natural number (whose type is abbreviated above as the range type [1..∞)) separate from the collected tuple of subsequent integers.

All that remains for the body is to sort by a custom comparator. Here we escape the need for a second-level comparison between x and -x by shifting the values slightly towards -∞.

Answer 30

dc, 53 bytes

Fd^sr[d*v]sA[ddlr+0=Asr]sR[dlAxlrlAx!<Rs_z0<M]dsMxlrp

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We're going to use the register r to hold our final result, so the first thing we need to do is put a too-big number in there so that our first comparison always succeeds. I think the largest number we can push in two bytes is 165 (which is less than the required 5526), but Fd^ gives us 15^15 which is plenty large. Macro A, [d*v]sA just does the square root of the square to give us absolute value.

Macro R handles putting a value into register r; we'll come back to it in a second. Macro M, [dlAxlrlAx!<Rs_z0<M]dsMx is our main macro. Duplicates the value on the stack, runs A to get ABS, then pushes the value in r and does the same. !<R checks to see if ABS(r) is greater than or equal to not less than ABS(top of stack), and if so it runs R. Since we had to leave behind a copy of the original top of stack for R to potentially use, we store to a useless register (s_) to pop it. z0<M just loops M until the stack is empty.

Macro R, [ddlr+0=Asr]sR duplicates the top of stack twice (we need to leave something behind for M to gobble up when we return). At this point we know that of the two absolute values, our new value is less than or equal to our old value. If they're equal, and one is positive and the other is negative, we always need the positive. Fortunately, knowing this we can easily check for the case of one being the inverse of the other by simply adding them and comparing to 0. If so, we run A, and no matter what else happened we know we're storing whatever is left in r.

After M has gone through the whole stack, we load r and print it.