2020.09.21 problem of the week

asteroids.py

# Given an array of integers representing asteroids in a row, for each
# asteroid, the absolute value represents its size, and the sign represents its
# direction (positive = right, negative = left). Return the state of the
# asteroids after all collisions (assuming they are moving at the same speed).
# If two asteroids meet, the smaller one will explode. When they are the same
# size, they both explode. Asteroids moving in the same direction will never
# meet.


def asteroids(a):
    # if there is to be a collision
    for i in range(len(a)):
        if len(a) >= i + 2 and a[i] >= 0 and a[i + 1] < 0:
            # mutual destruction
            if a[i] == abs(a[i + 1]):
                return asteroids(a[:i] + a[i + 2 :])
            # right moving asteroid wins
            elif a[i] > abs(a[i + 1]):
                return asteroids(a[: i + 1] + a[i + 2 :])
            # left moving asteroid wins
            else:
                return asteroids(a[:i] + a[i + 1 :])
    return a


assert asteroids([10, 9, 8, 7, 6, 5, 4, 3, 2, 1, -11]) == [-11]
assert asteroids([10, 9, -11]) == [-11]
assert asteroids([10, -10]) == []
assert asteroids([5, 8, -5]) == [
    5,
    8,
]  # The 8 and -5 collide, 8 wins. The 5 and 8 never collide.