CODEWITHSUNDEEP
rubyarraysmax
unj2
unj2

Reputation: 53481

Is there a better way to find the location of a minimum element in an Array?

Right now I have

def min(array,starting,ending)
  minimum = starting
  for i in starting+1 ..ending
    if array[i]<array[minimum]
      minimum = i
    end    
  end

return minimum
end

Is there a better "implementation" in Ruby? This one still looks c-ish. Thanks.

Upvotes: 3

Views: 1335

Answers (5)

Prajjwal
Prajjwal

Reputation: 1075

There is a simpler way and it works for me in ruby 1.9.2:

a = [6, 9, 5, 3, 0, 6]
a.find_index a.min

Upvotes: 1

rampion
rampion

Reputation: 89053

If you want to find the index of the minimal element, you can use Enumerable#enum_for to get an array of items-index pairs, and find the minimum of those with Enumerable#min (which will also be the minimum of the original array).

% irb
irb> require 'enumerator'
#=> true
irb> array = %w{ the quick brown fox jumped over the lazy dog }
#=> ["the", "quick", "brown", "fox", "jumped", "over", "the", "lazy", "dog"]
irb> array.enum_for(:each_with_index).min
#=> ["brown", 2]

If you want to bound it to specific array indices:

irb> start = 3
#=> 3
irb> stop = 7
#=> 7
irb> array[start..stop].enum_for(:each_with_index).min
#=> ["fox", 0]
irb> array[start..stop].enum_for(:each_with_index).min.last + start
#=> 3

Upvotes: 6

Ana Betts
Ana Betts

Reputation: 74654

Basically that's the best you can do, though you can write it a bit more succinctly:

def minval(arr)
    arr.inject {|acc,x| (acc && acc < x ? acc : x)}
end

Upvotes: 1

Bryan M.
Bryan M.

Reputation: 17322

If this isn't simply an academic question, why not just use Ruby's native sort method? It's implemented using a quicksort algorithm, and is considered to be pretty fast.

a = [3, 4, 5, 1, 7, 5]
a.sort![0] # => 1

Upvotes: -1

AlbertoPL
AlbertoPL

Reputation: 11509

This is the standard algorithm for finding the minimum element in an array, it can be better by having the array already be sorted before this function is called.

Otherwise I can't find a more efficient way of doing this. Specifically, linear time in big O notation is the best we can do.

Upvotes: 0

Related Questions