'Ruby - There's an array of integers given. I have to find the pair of adjacent elements that has the largest product

This little thing in the arcade mode of codesignal.com is literally killing me.

Here's the problem-

Given an array of integers, find the pair of adjacent elements that has the largest product and return that product. For inputArray = [3, 6, -2, -5, 7, 3], the output should be adjacentElementsProduct(inputArray) = 21. 7 and 3 produce the largest product.

This is the code I could manage:

def adjacentElementsProduct(inputArray)
    for check in 0..(inputArray.length-1)
        previous = -5000000000000000000000000000000       
        if check < inputArray.length-1
            name = inputArray[check] * inputArray[check+1]
            chosen = name if name > previous
        end
    end
    return chosen
end

This gives 3/9 correct answers.

arraysrubyfunction


Solution 1:[1]

def max_product_adjacent_elements(arr)
  arr.each_cons(2).max_by { |x,y| x*y }.reduce(:*)
end

arr = (-50..100).to_a.sample(10)
  #=> [100, 0, 4, 20, -45, 71, 21, 39, 40, -33]

max_product_adjacent_elements(arr)
  #=> 1560 (39*40)

See Enumerable#each_cons and Enumerable#max_by.

Here's another way.

mx = -Float::INFINITY
enum = arr.to_enum
  #=> #<Enumerator: [63, 76, 31, 25, 38, 91, 87, 34, 86, 72]:each>
loop { mx = [mx, enum.next * enum.peek].max }
mx
  #=> 1560

When the enumerator is at its last position in arr, Enumerator#peek raises a StopIteration exception that Kernel#loop handles by breaking out of the loop.

Solution 2:[2]

Just in case somebody would consider first and last elements as adjacent:

def max_product_adjacent_elements_circular(arr)
  prod = 0
  arr.size.times do
    new_prod = arr.rotate![0..1].reduce(:*)
    prod = new_prod if new_prod > prod
  end
  prod
end

Solution 3:[3]

arr.each_cons(2).max_by{|x, y| x*y }.inject(:*)

  • each_cons(N) function takes the block which is used to check the condition and N which specifies the number of consecutive elements to take.
  • follow by max which will return the maximum pair outcome
  • inject(:*) multiply the two

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1
Solution 2 iGian
Solution 3

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