RogerBW's Blog

I’ve been doing the Weekly Challenges. The latest involved searching in lists of integers. (Note that this closes today.)

Task 1: Kill and Win

You are given a list of integers.

Write a script to get the maximum points. You are allowed to take out (kill) any integer and remove from the list. However if you do that then all integers exactly one-less or one-more would also be removed. Find out the total of integers removed.

This is simpler than it looks, because the rules ensure that all numbers will eventually be removed, either by direct choice or by being adjacent to a direct choice - and both ways count towards the score. Therefore, whatever order this happens in, the actual answer is simply the sum of all the list values.

Raku:

sub killandwin(@a) {
    return @a.sum;
}

Task 2: Number Collision

You are given an array of integers which can move in right direction if it is positive and left direction when negative. If two numbers collide then the smaller one will explode. And if both are same then they both explode. We take the absolute value in consideration when comparing.

All numbers move at the same speed, therefore any 2 numbers moving in the same direction will never collide.

Write a script to find out who survives the collision.

I don't think there are any timing considerations - if elements 2 and/or 3 eliminate each other, their neighbours needn't be reconsidered until the next pass. Standard clean/dirty flag loop.

Ruby:

def numbercollision(aa)
  a = aa
  dirty = true
  while dirty do
    b = []
    dirty = false
    i = 0
    while i + 1 <= a.length do

If there's a collision at all:

      if i + 1 < a.length && a[i] > 0 && a[i + 1] < 0 then
        ab = a[i + 1].abs

Larger first number is retained

        if a[i] > ab then
          b.push(a[i])

Or larger second number is retained

        elsif a[i] < ab then
          b.push(a[i + 1])
        end

Or nothing is, if they were equal.

        i += 2
        dirty = true

Or there wasn't a collision and we just copy.

      else
        b.push(a[i])
        i += 1
      end
    end
    a = b;
  end
  return a
end

Full code on github.