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

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 2

Haskell, 63 43 bytes

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

Thanks to @MartinEnder for 4 bytes!

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

Pyth, 12 11 10 bytes

eSsM/#0,Vt

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

Answer 5

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 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

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 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

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 11

Jelly, 6 5 bytes

SfµƝṀ

Try it online!

-1 remembering it's not a dyadic chain

  µƝ     For each pair of adjacent elements, monadically:
S        Sum the pair,
 f       and filter it to only element[s] in the pair.
    Ṁ    Take the maximum.

Answer 12

Nekomata, 7 bytes

qᵗZđ+aṀ

Attempt This Online!

qᵗZđ+aṀ
q           Non-deterministically choose a contiguous subsequence
 ᵗZ         Check that it contains a 0
   đ        Unpair; check that its length is 2, and push the two elements
    +       Add
     a      All possible values
      Ṁ     Maximum

Answer 13

Jelly, 8 bytes

ṡ2ẠÐḟS€Ṁ

Try it online!

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

Answer 14

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 15

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 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

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 18

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 19

K (ngn/k), 14 bytes

|/+/'(*/')_2':

Try it online!

|/+/'(*/')_2':
           2':   pairs
     (*/')_      discard those whose product are nonzero
  +/'            sum of each pair
|/               maximum

Answer 20

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 21

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 22

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 23

Vyxal G, 36 bits^v2, 4.5 bytes

2lƛ∑w↔

Try it Online!

Porting Jelly is (still) always the best option

Explained

zṠ↔G­⁡​‎‎⁡⁠⁡‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁢‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁣‏‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁤‏‏​⁡⁠⁡‌­
z     # ‎⁡Overlapping pairs (equivalent to 2l)
 Ṡ    # ‎⁢summed (vectorised)
  ↔   # ‎⁣with only items in the input kept.
   G  # ‎⁤Get the biggest number in that list.
💎

Created with the help of Luminespire.

Answer 24

Nibbles, 8.5 bytes

`/.|!$>>@:?$0+$]

Attempt This Online!

Disgustingly long…

`/.|!$>>@:?$0+$]
    !    :       Zip
     $            the input
      >>          with the tail of
        @         the input
   |             Filter by
          ?$0     containing a zero
  .              Map
             +$   sum
`/             ] Maximum

Answer 25

BQN, 14 bytes

⌈´((«∨»)0⊸=)⊸/

Try

Seemed like a good excuse to use the shifts in «∨».

Answer 26

J, 16 bytes

}.>./@(+#~0=*)}:

Attempt This Online!

}.>./@(+#~0=*)}:
}.            }:    Input without first and without last, respectively
          0=*       elementwise product equals 0?
       +            elementwise sum
        #~          filter the left by the right being true
                    (the right is true iff one of the two is zero,
                    and the corresponding sum is the other number)
  >./@              maximum of those

Answer 27

Vyxal 3 G, 4 bytes

oᶻΠf

Try it Online!

oᶻΠf­⁡​‎‎⁡⁠⁡‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁢‏⁠‎⁡⁠⁣‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁤‏‏
o     # ‎⁡overlapping pairs
 ᶻΠ   # ‎⁢where the product is zero
   f  # flatten
💎

Created with the help of Luminespire.

Answer 28

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 29

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 30

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