Max-product-of-three.rb

#A non-empty zero-indexed array A consisting of N integers is given. The product of triplet (P, Q, R) equates to A[P] * A[Q] * A[R] (0 ≤ P < Q < R < N).
#For example, array A such that:
#  A[0] = -3
#  A[1] = 1
#  A[2] = 2
#  A[3] = -2
#  A[4] = 5
#  A[5] = 6
#contains the following example triplets:
#(0, 1, 2), product is −3 * 1 * 2 = −6
#(1, 2, 4), product is 1 * 2 * 5 = 10
#(2, 4, 5), product is 2 * 5 * 6 = 60
#Your goal is to find the maximal product of any triplet.
#Write a function:
#def solution(a)
#that, given a non-empty zero-indexed array A, returns the value of the maximal product of any triplet.
#For example, given array A such that:
#  A[0] = -3
#  A[1] = 1
#  A[2] = 2
#  A[3] = -2
#  A[4] = 5
#  A[5] = 6
#the function should return 60, as the product of triplet (2, 4, 5) is maximal.
#Assume that:
#N is an integer within the range [3..100,000];
#each element of array A is an integer within the range [−1,000..1,000].
#Complexity:
#expected worst-case time complexity is O(N*log(N));
#expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).
#Elements of input arrays can be modified.


def solution(a)
  return 0 if a.length < 3

  a.sort!

  right = a[a.length - 1] * a[a.length - 2] * a[a.length - 3]
  left = a[0] * a[1] * a[a.length - 1]

  return right > left ? right : left
end

p solution([-3, 1, 2, -2, 5, 6])
p solution([-3, 1, 2])
p solution([-9, 1, 2, 2])
p solution(( (-10..10).to_a * 2) + [-1000, 500, -1])

# grade:
# https://codility.com/demo/results/demoCDCGN7-259/