11.rb

# Brute force
# Times out
# O(n^2) time, O(1) space
def max_area(heights)
  max = 0
  current_left_index = nil
  current_right_index = nil
  
  heights.each_with_index do |num1, index1|
    next if num1 == 0
    # Assuming we can find another pole with num1 height, how far out do we need to go for it to be worth it?
    min_distance_2 = [index1 + 1, max / num1].max
    
    heights[min_distance_2..-1]&.each_with_index do |num2, index2|
      distance = min_distance_2 + index2 - index1
      height = [num1, num2].min
      
      if height * distance > max
        max = height * distance
      end
      max = [max, height * distance].max
    end
  end
  
  max
end

# 2 pointers
# O(N) time, O(1) space

def max_area(heights)
  pointer_a = 0
  pointer_b = heights.length - 1
  max_area = 0
  
  while pointer_a != pointer_b
    height_a = heights[pointer_a]
    height_b = heights[pointer_b]
    distance = pointer_b - pointer_a
    
    area = distance * [height_a, height_b].min
    max_area = [max_area, area].max
    
    if height_a < height_b
      pointer_a += 1
    else
      pointer_b -= 1
    end
  end
  
  max_area
end