Approach #1: Stack [Accepted]
Intuition
A row of asteroids is stable if no further collisions will occur. After adding a new asteroid to the right, some more collisions may happen before it becomes stable again, and all of those collisions (if they happen) must occur right to left. This is the perfect situation for using a stack.
Algorithm
Say we have our answer as a stack with rightmost asteroid top
, and a new
asteroid comes in. If new
is moving right (new > 0
), or if top
is moving left (top < 0
), no collision occurs.
Otherwise, if abs(new) < abs(top)
, then the new
asteroid will blow up; if abs(new) == abs(top)
then both asteroids will blow up; and if abs(new) > abs(top)
, then the top
asteroid will blow up (and possibly more asteroids will, so we should continue checking.)
Complexity Analysis

Time Complexity: , where is the number of asteroids. Our stack pushes and pops each asteroid at most once.

Space Complexity: , the size of
ans
.
Analysis written by: @awice.