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LC_Asteriod_Collision.py
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LC_Asteriod_Collision.py
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"""
735. Asteroid Collision
We are given an array asteroids of integers representing asteroids in a row.
For each asteroid, the absolute value represents its size, and the sign represents its direction (positive meaning right, negative meaning left). Each asteroid moves at the same speed.
Find out the state of the asteroids after all collisions. If two asteroids meet, the smaller one will explode. If both are the same size, both will explode. Two asteroids moving in the same direction will never meet.
Example 1:
Input: asteroids = [5,10,-5]
Output: [5,10]
Explanation: The 10 and -5 collide resulting in 10. The 5 and 10 never collide.
Example 2:
Input: asteroids = [8,-8]
Output: []
Explanation: The 8 and -8 collide exploding each other.
Example 3:
Input: asteroids = [10,2,-5]
Output: [10]
Explanation: The 2 and -5 collide resulting in -5. The 10 and -5 collide resulting in 10.
Constraints:
2 <= asteroids.length <= 104
-1000 <= asteroids[i] <= 1000
asteroids[i] != 0
"""
class Solution:
def asteroidCollision(self, asteroids: List[int]) -> List[int]:
ans = []
for new in asteroids:
# Handle the empty ans or left going asteroids keep adding
if len(ans) == 0 or ans[-1] < 0:
ans.append(new)
continue
# Right going asteriods plus more right going ones
if ans[-1] > 0 and new > 0:
ans.append(new)
continue
# Handle the right going one with a left going one
while(ans and ans[-1] > 0 and new < 0):
# If this is hit the new asteriod will be destroed
if ans[-1] > -new:
new = float('inf')
continue
# The current asteriod will be destroyed
if ans[-1] < -new:
ans.pop()
continue
# both the current and new asteriod will be destroyed
elif ans[-1] == -new:
ans.pop()
new = float('inf')
break
# Handle rest
if new != float('inf'):
ans.append(new)
return ans