Given that the spoiler time is finished (I think) these are my
solutions to the Quiz #131. I only worked on single arrays (and not
matrixes). If found 3 different solutions. The first one just searches
through the solution space and calculates the value for each subarray.
It takes O(n^3). The second one uses that the sum is associative to
use the previous calculation for the sum of a subarray to decrese the
time complexity to O(n^2). And, in the end, I’ve found a way to find
the max sub array in a single pass O(n). All the solutions have
constant space constraints (O(1)).
Here is the code (also at Parked at Loopia):
#!/usr/bin/env ruby
class Array
Runs in O(n^3), change the value function and find using any
objective function.
def max_subarray_original
better_subarray = []
better_value = 0
(0...self.length).each do |start|
(start...self.length).each do |finish|
value = value(start, finish)
if (value > better_value) then
better_value = value
better_subarray = self[start..finish]
end
end
end
better_subarray
end
def value(start, finish)
self[start…finish].inject(0) { |acum, value| acum+value }
end
Runs in O(n^2), uses the sum asociativity to avoid an iteration
through the array.
def max_subarray_optimized
better_subarray = []
better_value = 0
(0...self.length).each do |start|
value = 0
(start...self.length).each do |finish|
value += self[finish]
if (value > better_value) then
better_value = value
better_subarray = self[start..finish]
end
end
end
better_subarray
end
It’s technically imposible to improve it in time or space
complexity.
Runs in O(n) time and O(1) space*.
* Assumes that each number takes the same space in memory and that
additions, substractions and comparisions take constant time.
def max_subarray_single_pass
sum = 0
min_pos = -1
min_value = 0
min_pos_at_left = -1
min_value_at_left = 0
better_end_pos = -1
better_value = 0
self.each_with_index do
|value, index|
sum += value
if sum - min_value > better_value then
better_value = sum - min_value
better_end_pos = index
min_value_at_left = min_value
min_pos_at_left = min_pos
end
if sum < min_value then
min_value = sum
min_pos = index
end
end
return [] if better_end_pos == -1
return self[(min_pos+1)..better_end_pos]
end
end
some manual testing
[[-1, 2, 5, -1, 3, -2, 1],
[1, -1000, 100],
[-3, -2, -1]].each do
|array|
puts “array”
p array
puts “max_subarray_original”
p array.max_subarray_original
puts “max_subarray_optimized”
p array.max_subarray_optimized
puts “max_subarray_single_pass”
p array.max_subarray_single_pass
end
