Maximum sub-array

Question

Define the "maximum sub-array" of a given array as "a (consecutive) sub-array that has the biggest sum". Note there's no "non-zero" requirement. Output that sum.

Give a description of your code if possible.

Sample input 1:

1 2 3 -4 -5 6 7 -8 9 10 -11 -12 -13 14

Sample output 1: 24

Description 1:
The biggest sum is yielded by cutting 6 7 -8 9 10 out and summing up.

Sample input 2: -1 -2 -3
Sample output 2: 0
Description 2: It's simple :) An empty subarray is the "biggest".

Requirement:

• Don't read anything except stdin, and output should go to stdout.
• Standard loopholes restrictions apply.

Ranking: The shortest program wins this code-golf.

Answer 1

Husk, 6 4 bytes

▲ṁ∫ṫ

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      -- implicit input (list) xs  - eg. [-1,2,3]
   ṫ  -- get all tails of xs       -     [[-1,2,3],[2,3],[3],[]]
 ṁ∫   -- map & concat cumsum       -     [0,-1,1,4,0,2,5,0,3,0]
▲     -- get maximum               -     5

Answer 2

Python 3, 61 bytes

a=b=0
for x in eval(input()):a=max(x,a+x);b=max(a,b)
print(b)

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Algorithm stolen from Wikipedia.

Answer 3

Jelly, 6 bytes

ẆS€;0Ṁ

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Answer 4

Pyth, 8 bytes

eS+0sM.:

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---

How?

eS+0sM.:Q - Q is implicit, meaning input. Let's say it's [-1, -2, -3].

      .:  - All contiguous non-empty sublists. We have [[-1], [-2], [-3], [-1, -2], [-2, -3], [-1, -2, -3]].
    sM    - Get the sum of each sublist. [-1, -2, -3, -3, -5, -6]
  +0      - Append a 0 to the sum list. [0, -1, -2, -3, -3, -5, -6]
eS        - Maximum element. S gives us [-6, -5, -3, -3, -2, -1, 0], while e returns 0, the last element.

Answer 5

05AB1E, 4 bytes

Ό0M

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-1 thanks to Adnan.

Answer 6

Haskell, 28 bytes

maximum.scanl((max<*>).(+))0

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Answer 7

C++, 197 195 187 bytes

-10 bytes thanks to Zacharý

#include<vector>
#include<numeric>
int f(std::vector<int>v){int i=0,j,t,r=0;for(;i<v.size();++i)for(j=i;j<v.size();++j){t=std::accumulate(v.begin()+i,v.begin()+j,0);if(t>r)r=t;}return r;}

Answer 8

R, 54 bytes

a=b=0;for(x in scan()){a=max(x,a+x);b=max(a,b)};cat(b)

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Algorithm taken from: Maximum subarray problem

R, 65 bytes

y=seq(x<-scan());m=0;for(i in y)for(j in y)m=max(m,sum(x[i:j]));m

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• Read x from stdin.
• Set y as index of x.
• Iterate twice over all possible nonempty subsets.
• Compare sum of a subset with m (initially m=0).
• Store maximum value in m.
• Print value of m.

R, 72 bytes

n=length(x<-scan());m=0;for(i in 1:n)for(j in i:n)m=max(m,sum(x[i:j]));m

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• Read x from stdin.
• Do a full search over all possible nonempty subsets.
• Compare sum of a subset with m (initially m=0).
• Store maximum value in m.
• Print value of m.

Other unsuccessful ideas

58 bytes

Reduce(max,lapply(lapply(seq(x<-scan()),tail,x=x),cumsum))

63 bytes

Reduce(max,lapply(seq(x<-scan()),function(i)cumsum(tail(x,i))))

72 bytes

m=matrix(x<-scan(),n<-length(x),n);max(apply(m*lower.tri(m,T),2,cumsum))

Answer 9

05AB1E, 4 bytes

-2 bytes thanks to Adnan

ÎŒOM

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Answer 10

Mathematica, 24 bytes

Max[Tr/@Subsequences@#]&

Answer 11

Haskell, 41 33 bytes

import Data.List
g=maximum.concatMap(map sum.inits).tails

maximum.(scanl(+)0=<<).scanr(:)[]

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Answer 12

Gaia, 6 bytes

0+ḋΣ¦⌉

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Answer 13

k, 14 bytes

|/,/+\'(1_)\0,

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            0, /prepend a zero (in case we're given all negatives)
       (1_)\   /repeatedly remove the first element, saving each result
    +\'        /cumulative sum over each result, saving each result
  ,/           /flatten (fold concat)
|/             /maximum (fold max)

Answer 14

Japt, 11 bytes

£ãY mxÃc rw

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Explanation

£ãY mxÃc rw
m@ãY mx} c rw   // Ungolfed
m@     }        // Map the input array by the following function, with Y=index
  ãY            //   Get all subsections in input array length Y
     mx         //   Sum each subsection
         c rw   // Flatten and get max

Alternate method, 11 bytes

From @ETHproductions; based on Brute Forces' Husk answer.

£sY å+Ãc rw

Gets all tails of the input array and cumulatively sums each. Then flattens the array and gets the max.

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Answer 15

Ruby, 61 59 57 bytes

I just started learning Ruby, so this is what I came up with.

s=0
p [gets.split.map{|i|s=[j=i.to_i,s+j].max}.max,0].max

I first saw this algorithm at the Finnish version of this unfinished book. It is very well explained at the page 23.

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Answer 16

JavaScript, 58 bytes

m=Math.max;x=y=>eval("a=b=0;for(k of y)b=m(a=m(a+k,k),b)")

Golfed JS implementation of Kadane's algorithm. Made as short as possible. Open to constructive suggestions!

What I learnt from this post: return value of eval - when its last statment is a for loop - is basically the last value present inside the loop. Cool!

EDIT: saved four bytes thanks to Justin's and Hermann's suggestions.

Answer 17

J, 12 bytes

[:>./@,+/\\.

Similar to zgrep's K solution: the scan sum of all suffixes (produces a matrix), raze, take max

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NOTE

for not too many more bytes, you can get an efficient solution (19 bytes golfed):

[: >./ [: ({: - <./)\ +/\

Answer 18

Japt -h, 4 bytes

ãx ñ

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ãx ñ     :Implicit input of array
ã        :Sub-arrays
 x       :Reduced by addition
   ñ     :Sort
         :Implicit output of last element

Answer 19

Python 2, 52 51 bytes

f=lambda l:len(l)and max(sum(l),f(l[1:]),f(l[:-1]))

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Answer 20

Common Lisp, 73 bytes

(lambda(a &aux(h 0)(s 0))(dolist(x a s)(setf h(max x(+ h x))s(max s h))))

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Answer 21

APL, 31 29 27 bytes

⌈/∊∘.{+/W[X/⍨⍺≤X←⍳⍵]}⍨⍳⍴W←⎕

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How?

• ∊∘.{+/W[X/⍨⍺≤X←⍳⍵]}⍨⍳⍴W←⎕ Generate sums of subvectors
• ⌈/ Maximum

Answer 22

CJam, 24 bytes

q~:A,{)Aew{:+}%}%e_0+:e>

Function that takes array of numbers as input.

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q~:A   e# Store array in 'A' variable
,{)Aew e# Get every possible sub-array of the array
{:+}%  e# Sum every sub array
}e_    e# flatten array of sums
0+     e# Add zero to the array
:e>    e# Return max value in array

Answer 23

MY, 11 bytes

⎕𝟚35ǵ'ƒ⇹(⍐↵

Try it online! MY is on TIO now! Woohoo!

How?

• ⎕ = evaluated input
• 𝟚 = subvectors
• 35ǵ' = chr(0x53) (Σ, sum)
• ƒ = string as a MY function
• ⇹ = map
• ( = apply
• ⍐ = maximum
• ↵ = output with a newline.

Sum was fixed (0 on empty arrays) in order for this to work. Product was also fixed.

Answer 24

Axiom, 127 bytes

f(a:List INT):Complex INT==(n:=#a;n=0=>%i;r:=a.1;for i in 1..n repeat for j in i..n repeat(b:=reduce(+,a(i..j));b>r=>(r:=b));r)

This would be O(#a^3) Algo; I copy it from the C++ one... results

(3) -> f([1,2,3,-4,-5,6,7,-8,9,10,-11,-12,-13,14])
   (3)  24
                                                    Type: Complex Integer
(4) -> f([])
   (4)  %i
                                                    Type: Complex Integer
(5) -> f([-1,-2,3])
   (5)  3
                                                    Type: Complex Integer

Answer 25

Scala, 105 bytes

val l=readLine.split(" ").map(_.toInt);print({for{b<-l.indices;a<-0 to b+2}yield l.slice(a,b+1).sum}.max)

I didn't find any better way to generate the sublists arrays.

Answer 26

Java 8, 242 bytes

import java.util.*;v->{List a=new Stack();for(String x:new Scanner(System.in).nextLine().split(" "))a.add(new Long(x));int r=0,l=a.size(),i=l,j,k,s;for(;i-->0;)for(j=l;--j>1;r=s>r?s:r)for(s=0,k=i;k<j;)s+=(long)a.get(k++);System.out.print(r);}

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106 bytes without using STDIN/STDOUT requirement.. >.>

a->{int r=0,l=a.length,i=l,j,k,s;for(;i-->0;)for(j=l;--j>1;r=s>r?s:r)for(s=0,k=i;k<j;s+=a[k++]);return r;}

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Explanation:

import java.util.*;      // Required import for List, Stack and Scanner

v->{                     // Method with empty unused parameter and no return-type
  List a=new Stack();    //  Create a List
  for(String x:new Scanner(System.in).nextLine().split(" "))
                         //  Loop (1) over the STDIN split by spaces as Strings
    a.add(new Long(x));  //   Add the String converted to a number to the List
                         //  End of loop (1) (implicit / single-line body)
  int r=0,               //  Result-integer
      l=a.size(),        //  Size of the List
      i=l,j,k,           //  Index-integers
      s;                 //  Temp sum-integer
  for(;i-->0;)           //  Loop (2) from `l` down to 0 (inclusive)
    for(j=l;--j>1;       //   Inner loop (3) from `l-1` down to 1 (inclusive)
        r=               //     After every iteration: change `r` to:
          s>r?           //      If the temp-sum is larger than the current `r`:
           s             //       Set `r` to the temp-sum
          :              //      Else:
           r)            //       Leave `r` the same
      for(s=0,           //    Reset the temp-sum to 0
          k=i;k<j;)      //    Inner loop (4) from `i` to `j` (exclusive)
        s+=(long)a.get(k++);
                         //     Add the number at index `k` in the List to this temp-sum
                         //    End of inner loop (4) (implicit / single-line body)
                         //   End of inner loop (3) (implicit / single-line body)
                         //  End of loop (2) (implicit / single-line body)
  System.out.print(r);   //  Print the result to STDOUT
}                        // End of method