Find the largest number that's adjacent to a zero

Question

Challenge:

Take a vector / list of integers as input, and output the largest number that's adjacent to a zero.

Specifications:

• As always, optional input and output format
• You may assume that there will be at least one zero, and at least one non-zero element.

Test cases:

1 4 3 6 0 3 7 0
7

9 4 9 0 9 0 9 15 -2
9

-4 -6 -2 0 -9
-2

-11 0 0 0 0 0 -12 10
0

0 20 
20

Good luck and happy golfing!

Answer 1

Jelly, 8 bytes

ṡ2ẠÐḟS€Ṁ

Try it online!

ṡ2            Overlapping pairs
  ẠÐḟ         Remove pairs without zeroes
     S€       Sum of each pair
       Ṁ      Maximum

Answer 2

MATL, 10 bytes

t~5BZ+g)X>

Try it online! Or verify all test cases.

Explanation

Let's take input [-4 -6 -2 0 -9] as an example.

t     % Input array. Duplicate
      %   STACK: [-4 -6 -2 0 -9],  [-4 -6 -2 0 -9]
~     % Logical negate. Replaces zeros by logical 1, and nonzeros by logical 0
      %   STACK: [-4 -6 -2 0 -9],  [0 0 0 1 0]
5B    % Push logical array [1 0 1] (5 in binary)
      %   STACK: [-4 -6 -2 0 -9], [0 0 0 1 0], [1 0 1]
Z+    % Convolution, maintaining size. Gives nonzero (1 or 2) for neighbours of
      % zeros in the original array, and zero for the rest
      %   STACK: [-4 -6 -2 0 -9], [0 0 1 0 1]
g     % Convert to logical
      %   STACK: [-4 -6 -2 0 -9], [0 0 1 0 1]
)     % Use as index into original array
      %   STACK: [-2 -9]
X>    % Maximum of array.
      %   STACK: -2
      % Implicitly display

Answer 3

05AB1E, 9 bytes

ü‚D€P_ÏOZ

Explanation

ü‚         # pair up elements
  D        # duplicate
   €P      # product of each pair (0 if the pair contains a 0)
     _     # logical negate, turns 0 into 1 and everything else to 0
      Ï    # keep only the pairs containing at least 1 zero
       O   # sum the pairs
        Z  # take max

Doesn't work in the online interpreter, but works offline.

Answer 4

Haskell, 63 43 bytes

f x=maximum[a+b|(a,b)<-tail>>=zip$x,a*b==0]

Thanks to @MartinEnder for 4 bytes!

Answer 5

Pyth, 12 11 10 bytes

eSsM/#0,Vt

Forms pairs, filters by zero member, sorts by sum, returns largest.

Answer 6

JavaScript (ES6), 59 57 56 bytes

let f =
    
l=>l.map((n,i)=>m=l[i-1]==0|l[i+1]==0&&n>m?n:m,m=-1/0)|m

console.log(f([1, 4, 3, 6, 0, 3, 7, 0]));       // 7
console.log(f([9, 4, 9, 0, 9, 0, 9, 15, -2]));  // 9
console.log(f([-4, -6, -2, 0, -9]));            // -2
console.log(f([-11, 0, 0, 0, 0, 0, -12, 10]));  // 0
console.log(f([3, 0, 5]));                      // 5
console.log(f([28, 0, 14, 0]));                 // 28

Edit: saved 2 bytes thanks to Huntro
Edit: saved 1 byte thanks to ETHproductions

Answer 7

JavaScript (ES6), 53 bytes

a=>(m=-1/0,a.reduce((l,r)=>(m=l*r||l+r<m?m:l+r,r)),m)

Because I like using reduce. Alternative solution, also 53 bytes:

a=>Math.max(...a.map((e,i)=>e*a[++i]==0?e+a[i]:-1/0))

Answer 8

Python, 49 bytes

lambda a:max(sum(x)for x in zip(a,a[1:])if 0in x)

Tests are at ideone

Zips through the pairs, sums the ones containing any zero, returns the maximum.

Answer 9

PHP, 77 68 71 bytes

-3 bytes from anonymous, -4 and -2 from MartinEnder

preg_match_all("#(?<=\b0 )\S+|\S+(?= 0)#",$argv[1],$m);echo max($m[0]);

run with php -r '<code>' '<space separated values>'

Answer 10

Ruby, 51 bytes

->a{a.each_cons(2).map{|a,b|a*b!=0?-1.0/0:a+b}.max}

usage

f=->a{a.each_cons(2).map{|a,b|a*b!=0?-1.0/0:a+b}.max}
p f[gets.split.map(&:to_i)]

Answer 11

Java 7, 118 105 106 bytes

int d(int[]a){int i=0,m=1<<31,c;for(;++i<a.length;m=a[i]*a[i-1]==0&(c=a[i]+a[i-‌​1])>m?c:m);return m;}

13 bytes saved thanks to @cliffroot by using an arithmetic approach instead. 1 additional byte thank to @mrco after he discovered a bug (the added test case 2, 1, 0 would return 2 instead of 1).

Ungolfed & test code:

Try it here.

class M{
  static int c(int[] a){
    int i,
        m = a[i=0],
        c;
    for(; ++i < a.length; m = a[i] * a[i-1] == 0 & (c = a[i] + a[i - 1]) > m)
                           ? c
                           : m);
    return m;
  }

  public static void main(String[] a){
    System.out.println(c(new int[]{ 1, 4, 3, 6, 0, 3, 7, 0 }));
    System.out.println(c(new int[]{ 9, 4, 9, 0, 9, 0, 9, 15, -2 }));
    System.out.println(c(new int[]{ -4, -6, -2, 0, -9 }));
    System.out.println(c(new int[]{ -11, 0, 0, 0, 0, 0, -12, 10 }));
    System.out.println(c(new int[]{ 0, 20 }));
    System.out.println(c(new int[]{ 2, 1, 0 }));
  }
}

Output:

7
9
-2
0
20
1

Answer 12

CJam, 16 bytes

q~2ew{0&},::+:e>

Try it online! (As a test suite.)

Explanation

q~    e# Read and eval input.
2ew   e# Get all (overlapping) pairs of adjacent values.
{0&}, e# Keep only those that contain a 0.
::+   e# Sum each pair to get the other (usually non-zero) value.
:e>   e# Find the maximum.

Answer 13

MATLAB with Image Processing Toolbox, 32 bytes

@(x)max(x(imdilate(~x,[1 0 1])))

This is an anonymous function. Example use for the test cases:

>> f = @(x)max(x(imdilate(~x,[1 0 1])))
f =
  function_handle with value:
    @(x)max(x(imdilate(~x,[1,0,1])))

>> f([1 4 3 6 0 3 7 0])
ans =
     7

>> f([9 4 9 0 9 0 9 15 -2])
ans =
     9

>> f([-4 -6 -2 0 -9])
ans =
    -2

>> f([-11 0 0 0 0 0 -12 10])
ans =
     0

>> f([0 20])
ans =
    20

Answer 14

Mathematica, 46 43 bytes

^{Saved 3 bytes due to @MartinEnder.}

Max[Tr/@Partition[#,2,1]~Select~MemberQ@0]&

Anonymous function. Takes a list of integers as input and returns an integer as output. Based off of the Ruby solution.

Answer 15

R, 48 47 bytes

EDIT: Fixed an error thanks to @Vlo and changed it to read input from stdins, saved one byte by assigning w and skipping parantheses.

function(v)sort(v[c(w<-which(v==0)-1,w+1)],T)[1]

v=scan();w=which(v==0);sort(v[c(w-1,w+1)],T)[1]

Unnested explanation

1. Find indices for which the vector v takes on the values 0: w <- which(v == 0)
2. Create new vector which contains the indices +-1: w-1 and w+1
3. Extract elements that match the indices w-1 and w+1
4. Sort in descending order and extract fist element

Note that if the last or first element of v is a zero, w+-1 will effectively fetch an index outside of the length of the vector which implies that v[length(v)+1] returns NA. This is generally no problem but the max() functions inconveniently returns NA if there are any occurrences in the vector unless one specifies the option na.rm=T. Thus it is 2 bytes shorter to sort and extract first element than to use max(), e.g.:

max(x,na.rm=T)
sort(x,T)[1]

Answer 16

Dyalog APL, 14 bytes

⌈/∊2(+↑⍨0∊,)/⎕

⌈/ largest of

∊ the flattened ("enlisted"

2(...)/ pairwise

 + sum (zero plus something is something)

 ↑⍨ taken if

 0 zero

 ∊ is a member of

 , the pair (lit. the concatenation of the left-hand number and the right-hand number)

TryAPL online!

Answer 17

Julia, 56 55 Bytes

f(l)=max(map(sum,filter(t->0 in t,zip(l,l[2:end])))...)

Create tuples for neighboring values, take those tuples containing 0, sum tuple values and find maximum

Answer 18

Perl, 42 bytes

Includes +1 for -p

Give the numbers on line on STDIN

largest0.pl <<< "8 4 0 0 5 1 2 6 9 0 6"

largest0.pl:

#!/usr/bin/perl -p
($_)=sort{$b-$a}/(?<=\b0 )\S+|\S+(?= 0)/g

Answer 19

Python 2, 74 Bytes

def f(x):x=[9]+x;print max(x[i]for i in range(len(x)) if 0in x[i-1:i+2:2])

Cycle through every element, if there's a 0 in the position of either the left or the right of the current element, include it in the generator, and then run it through max. We need to pad the list with some non-0 number. It'll never be included because the slice [-1:2:2] won't include anything.

Answer 20

C# 76 74 bytes

using System.Linq;i=>i.Zip(i.Skip(1),(a,b)=>a*b==0?1<<31:a+b).Max(‌​);

Explanation:

Use zip to join the array with itself but skipping the first value in the 2nd reference so that item zero joins to item one. Multiply a times b, if the result is zero, one of them must be zero and output a + b. Otherwise, output the minimum possible integer in the language. Given the assumption that we will always have a zero and a non-zero, this minimum value will never be output as the max.

Usage:

[TestMethod]
public void LargestFriend()
{
    Assert.AreEqual(7, F(new int[] { 1, 4, 3, 6, 0, 3, 7, 0 }));
    Assert.AreEqual(9, F(new int[] { 9, 4, 9, 0, 9, 0, 9, 15, -2 }));
    Assert.AreEqual(-2, F(new int[] { -4, -6, -2, 0, -9 }));
    Assert.AreEqual(0, F(new int[] { -11, 0, 0, 0, 0, 0, -12, 10 }));
    Assert.AreEqual(20, F(new int[] { 0, 20 }));
}

Answer 21

T-SQL, 182 bytes

Golfed:

DECLARE @x varchar(max)='1 5 4 3 6 1 3 17 1 -8 0 -7'

DECLARE @a INT, @b INT, @ INT WHILE @x>''SELECT @a=@b,@b=LEFT(@x,z),@x=STUFF(@x,1,z,''),@=IIF(@a=0,IIF(@b<@,@,@b),IIF(@b<>0 or @>@a,@,@a))FROM(SELECT charindex(' ',@x+' ')z)z PRINT @

Ungolfed:

DECLARE @x varchar(max)='1 5 4 3 6 1 3 17 1 -8 0 -7'

DECLARE @a INT, @b INT, @ INT
WHILE @x>''
  SELECT 
   @a=@b,
   @b=LEFT(@x,z),
   @x=STUFF(@x,1,z,''),
   @=IIF(@a=0,IIF(@b<@,@,@b),IIF(@b<>0 or @>@a,@,@a))
  FROM(SELECT charindex(' ',@x+' ')z)z 
PRINT @

Fiddle

Answer 22

PowerShell v3+, 62 bytes

param($n)($n[(0..$n.count|?{0-in$n[$_-1],$n[$_+1]})]|sort)[-1]

A bit longer than the other answers, but a nifty approach.

Takes input $n. Then loops through the indices 0..$n.count, uses the Where-Object (|?{...}) to pull out those indices where the previous or next item in the array is 0, and feeds those back into array slice $n[...]. We then |sort those elements, and take the biggest [-1].

Examples

PS C:\Tools\Scripts\golfing> @(1,4,3,6,0,3,7,0),@(9,4,9,0,9,0,9,15,-2),@(-4,-6,-2,0,-9),@(-11,0,0,0,0,0,-12,10)|%{""+$_+" --> "+(.\largest-number-beside-a-zero.ps1 $_)}
1 4 3 6 0 3 7 0 --> 7
9 4 9 0 9 0 9 15 -2 --> 9
-4 -6 -2 0 -9 --> -2
-11 0 0 0 0 0 -12 10 --> 0

PS C:\Tools\Scripts\golfing> @(0,20),@(20,0),@(0,7,20),@(7,0,20),@(7,0,6,20),@(20,0,6)|%{""+$_+" --> "+(.\largest-number-beside-a-zero.ps1 $_)}
0 20 --> 20
20 0 --> 20
0 7 20 --> 7
7 0 20 --> 20
7 0 6 20 --> 7
20 0 6 --> 20

Answer 23

q, 38 bytes

{max x where 0 in'x,'(next x),'prev x}

Answer 24

J, 18 bytes

[:>./2(0&e.\#+/\)]

Explanation

[:>./2(0&e.\#+/\)]  Input: array A
                 ]  Identity. Get A
     2              The constant 2
      (         )   Operate on 2 (LHS) and A (RHS)
               \    Get each subarray of size 2 from A and
             +/       Reduce it using addition
           \        Get each subarray of size 2 from A and
       0&e.           Test if 0 is a member of it
            #       Filter for the sums where 0 is contained
[:>./               Reduce using max and return

Answer 25

Perl 6, 53 bytes

{max map ->$/ {$1 if !$0|!$2},(1,|@_,1).rotor(3=>-2)}

Expanded:

# bare block lambda with implicit signature of (*@_)
{
  max

    map

      -> $/ {           # pointy lambda with parameter ｢$/｣
                        # ( ｢$0｣ is the same as ｢$/[0]｣ )
        $1 if !$0 | !$2 # return the middle value if either of the others is false
      },

      ( 1, |@_, 1 )     # list of inputs, with added non-zero terminals
      .rotor( 3 => -2 ) # grab 3, back-up 2, repeat until less than 3 remain
}

Answer 26

PHP, 66 bytes

foreach($a=$argv as$k=>$v)$v||$m=max($m,$a[$k-1],$a[$k+1]);echo$m;

Pretty straightforward. Iterates over the input, and when a number is 0, it sets $m to the highest number of the 2 adjacent numbers and any previous value of $m.

Run like this (-d added for aesthetics only):

php -d error_reporting=30709 -r 'foreach($a=$argv as$k=>$v)$v||$m=max($m,$a[$k-1],$a[$k+1]);echo$m;' -- -4 -6 -2 0 -9;echo

Answer 27

R, 48 54 bytes

s=scan()

w=which;max(s[c(w(s==0)+1,w(s==0)-1)],na.rm=T)

Reads vector from console input, then takes the maximum over all values adjacent to 0.

Edit: Catches NAs produced at the boundary, thanks rturnbull!

Answer 28

Racket 183 bytes

(λ(k)(let*((lr(λ(l i)(list-ref l i)))(l(append(list 1)k(list 1)))(m(for/list((i(range 1(sub1(length l))))
#:when(or(= 0(lr l(sub1 i)))(= 0(lr l(add1 i)))))(lr l i))))(apply max m)))

Detailed version:

(define f
 (λ(k)
    (let* ((lr (λ(l i)(list-ref l i)))
           (l (append (list 1) k (list 1)))
           (m (for/list ((i (range 1 (sub1(length l))))
                         #:when (or (= 0 (lr l (sub1 i)))
                                    (= 0 (lr l (add1 i))) ))
                (lr l i) )))
      (apply max m) )))

Testing:

(f (list 1 4 3 6 0 3 7 0))
(f (list 9 4 9 0 9 0 9 15 -2))
(f (list -4 -6 -2 0 -9))
(f (list -11 0 0 0 0 0 -12 10))
(f (list 0 20 ))

Output:

7
9
-2
0
20

Answer 29

C 132 bytes

Outputs using main's return code:

int main(int a,char**_){int i,m=0;_[0]=_[a]="1";for(i=1;i<a;++i){m=(*_[i-1]-48||*_[i+1]-48?m>atoi(_[i])?m:atoi(_[i]):m);}return m;}

I feel like I should be able to save a few bytes by saving one of the atoi calls, but I couldn't find an efficient way. (,t plus t= plus , plus t twice is too long). Also this technically uses undefined behaviour (setting _[a] to "1") but every compiler I know of allows it by default.

Strategy: pad the start and end of the array with 1, then loop over the internal section checking each neighbor.

Answer 30

PHP 69 64 bytes

Some bytes on and off from Jörg Hülsermann and Titus. = (-5)

Requires register_globals enabled. Usage: http://localhost/notnull.php?i[]=9&i[]=-5i[]=...

$x=$_GET['i'];
$y=0;
foreach($x as $j){
if($y<abs($j)){
$y=$j;
}
}
echo $y;

Golfed:

$x=$_GET['i'];$y=0;foreach($x as $j)if($y<abs($j))$y=$j;echo $y;

Answer 31

SPSS Syntax (98 bytes)

Golfed solution:

CRE L=lead(A,1).
COMP R=MAX(lag(A),L).
EXE.
SEL IF A=0.
EXE.
AGG
/S=MAX(R).
LIST S
/CAS=1.

Ungolfed:

CREATE L=lead(A,1).
COMPUTE R=MAX(lag(A),L).
EXECUTE.
SELECT IF A=0.
EXECUTE.
AGGREGATE
/S=MAX(R).
LIST S
/CASES=1.

Explanation: Input values is ordered vertically in a column. The first line of code creates a column with the lead number, the number before any given number. MAX(lag,A) returns the lagging number, the number after any given number (in test series A). So the second line of code creates a column with the highest number of neighbourging numbers of any given number. We are only interested in the neighbours of zeros, so the forth line selects all the rows with zeros, which now also include the highest valued neighbours. The 6th and 7th line of code finds the max value of selected neighbours.

With data inputs:

* This syntax solves: Returning the greatest adjacent number bigger or equal to itself,
  for any number V.
* CodeGolf asks for the solution for the case V=0.
* Bytes=98.

DATA LIST LIST
/ Sample (A7) a b c d e f g h i j k l m .
BEGIN DATA.
"A" 1, 4, 3, 6, 0, 3, 7, 0
"B" 9, 4, 9, 0, 9, 0, 9, 15, -2
"C" -4, -6, -2, 0, -9
"D" -11, 0, 0, 0, 0, 0, -12, 10
"E"  0, 20
END DATA.
DATASET NAME Inputs WINDOW=FRONT.

FLIP VARIABLES=a b c d e f g h i
/NEWNAMES=Sample.
DATASET NAME CodeGolf WINDOW=FRONT.
DATASET CLOSE Inputs.

CRE L=lead(A,1).
COMP R=MAX(lag(A),L).
EXE.
SEL IF A=0.
EXE.
AGG
/S=MAX(R).
LIST S
/CAS=1.

Just change the sample references (A to E) for testing each sample.