Create asteroid-collision.py

Author: kamyu104

Python/asteroid-collision.py

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+# Time:  O(n)
+# Space: O(n)
+
+# 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.
+# Example 4:
+# Input: 
+# asteroids = [-2, -1, 1, 2]
+# Output: [-2, -1, 1, 2]
+# Explanation: 
+# The -2 and -1 are moving left, while the 1 and 2 are moving right.
+# Asteroids moving the same direction never meet, so no asteroids will meet each other.
+#
+# Note:
+# - The length of asteroids will be at most 10000.
+# - Each asteroid will be a non-zero integer in the range [-1000, 1000].
+
+class Solution(object):
+    def asteroidCollision(self, asteroids):
+        """
+        :type asteroids: List[int]
+        :rtype: List[int]
+        """
+        result = []
+        for asteroid in asteroids:
+            while result and asteroid < 0 < result[-1]:
+                if result[-1] < -asteroid:
+                    result.pop()
+                    continue
+                elif result[-1] == -asteroid:
+                    result.pop()
+                break
+            else:
+                result.append(asteroid)
+        return result