This is a classic algorithmic challenge and despite its seeming

simplicity,

there’s quite a bit you can learn from it. I’m pretty sure it was this

exact

problem that finally got Big O Notation through my thick skull, so I’ll

take

that approach with this summary.

The obvious code that tends to come to mind for solving this problem is

a

brute-force search through the subarrays. That’s not a bad thing. It’s

very

easy to code up and may work for you if the inputs are small enough. It

certainly works for the quiz example.

Here’s a solution of that by Drew O.:

class Array

# sum the integer values of array contents

def int_sum

self.inject(0){|sum,i| sum+i.to_i}

end

```
# find the maximum sub array in an array
def max_sub_array
(0...self.size).inject([self.first]) do |max_sub,i|
(i...self.size).each do |j|
if max_sub.int_sum < self[i..j].int_sum
max_sub = self[i..j]
end
end
max_sub
end
end
```

end

# test example

if **FILE** == $0

my_arr = [-1, 2, 5, -1, 3, -2, 1]

puts “array: #{my_arr.inspect}”

puts “maximum sub-array: #{my_arr.max_sub_array.inspect}”

end

I removed a little of Drew’s printing code in the above so we could

focus on the

algorithm, but the results are unchanged.

We can see that Drew’s code works by walking through all of the indices,

with

all possible lengths, to check each subarray. Each subarray is tested

against

the current best sum and the end result is that the highest total found

will be

returned.

The question we want to ask though, is how long does this take for

various

inputs? It’s quite zippy for the quiz example:

$ time ruby -r max_sub_array -e ‘p [-1, 2, 5, -1, 3, -2,

1].max_sub_array’

[2, 5, -1, 3]

real 0m0.013s

user 0m0.007s

sys 0m0.005s

It slows down pretty quick though, with bigger inputs:

$ time ruby -r max_sub_array

-e ‘p Array.new(100) { rand(11) - 5 }.max_sub_array’

[4, 0, 1, 4, -1, 1, 4, 0, 4, 3, 1, 0, 3, -4, 1, 4, -1, 0, 4, -3, 1,

-3, 4, 2]

real 0m0.307s

user 0m0.301s

sys 0m0.006s

$ time ruby -r max_sub_array

-e ‘p Array.new(1_000) { rand(11) - 5 }.max_sub_array’

[3, 1, -3, -1, 2, 5, 4, 3, -5, -2, 3, 1, 1, -2, -3, 4, 5, 4, 4, -3,

-1, â€¦]

real 3m39.856s

user 3m38.455s

sys 0m0.343s

The issue here is that those nested loops just execute many, many times.

In

fact, that inner each() is called 500,500 times for an Array with 1,000

entries.

If we want to tackle those bigger lists we need to lower that count.

One way to find the algorithms with lower iteration counts is to

randomly spot

check the solutions on an Array of similar length. In doing so, I

stumbled

across this solution from Justin Either:

$ time ruby -r max_sub_array

-e ‘p MaxSubArray.new.find(Array.new(1_000) { rand(11) - 5

})’

[1, 4, 1, 2, -5, -3, 4, 2, 3, 1, -2, 4, 5, 1, 3, 0, 5, -1, 4, 4, 2, 4,

â€¦]

real 0m0.016s

user 0m0.009s

sys 0m0.006s

$ time ruby -r max_sub_array

-e ‘p MaxSubArray.new.find(Array.new(10_000) { rand(11) -

5 })’

[3, 1, -2, 5, 4, 5, 0, -3, 0, 3, 5, -3, -4, -3, 5, -3, -1, 4, 5, -3,

3, â€¦]

real 0m0.047s

user 0m0.030s

sys 0m0.006s

$ time ruby -r max_sub_array

-e ‘p MaxSubArray.new.find(Array.new(100_000) { rand(11) -

5 })’

[4, 1, -2, 3, 4, -4, -4, 4, 5, 1, -3, 4, -5, 5, -5, 1, -1, 0, -5, 1,

-1, â€¦]

real 0m0.286s

user 0m0.267s

sys 0m0.011s

As you can see, that’s scaling to much higher counts much quicker.

That’s a

sure sign of a more clever algorithm, so let’s take a peek at the code:

# Object defining a sub-array of integer values

# The sub-array contains a start and end index

# defining a region of the master array

class SubArray

def initialize

@start = 0

@end = 0

@sum = 0

end

# Set boundaries of the sub-array

def set_bounds(list_start, list_end)

@start, @end = list_start, list_end

end

# Provide get/set accessors

attr_reader :start, :end, :sum

attr_writer :sum

end

class MaxSubArray

# Finds the sub-array with the largest sum

# Input: a list of integers

def find(list)

max = SubArray.new

cur = SubArray.new

```
for i in 0...list.size
cur.sum = cur.sum + list[i]
if (cur.sum > max.sum)
max.sum = cur.sum
cur.set_bounds(cur.start, i)
max.set_bounds(cur.start, i)
elsif (cur.sum < 0)
# If sum goes negative, this region cannot have the max sum
cur.sum = 0
cur.set_bounds(i + 1, i + 1)
end
end
list.slice(max.start, max.end - max.start + 1)
```

end

end

First, you need to take a look at the SubArray class. This is just a

bookkeeping tool to keep track of the bounds and sum of any given

subarray.

There is a minor bug here that hinders this solution on all-negative

Arrays like

[-3, -1], but it can be fixed by initializing sum to -1.0/0.0 instead of

0.

The second class holds the actual algorithm. The trick here is pretty

simple

once you’ve seen it before. Basically, we start from the beginning of

the Array

and expand the subarray to following indices. We keep track of the best

total

seen thus far and replace that when we find better totals.

The trick is that we hop our subarray indices forward whenever the

running total

dips into the negatives. A negative total is effectively starting over,

so

skipping over all of those numbers costs us nothing. The run is broken.

This algorithm in linear, so that for iterator only executes 1,000 times

for an

Array of that length. That’s where your big speed gain comes from in

this case

and the reason algorithms are important when dealing with larger inputs.

My thanks to all the algorists that showed off the variety of solutions

that can

be applied here.

Ruby Q. will now take a one week break to allow everyone the chance to

compete

in the ICFP Contest (http://www.icfpcontest.org/). If you are a fan of

programming contests, I strongly encourage you to give this yearly

competition a

shot. It’s always challenging and rewarding. Best of luck!