There are n 1-indexed robots, each having a position on a line, health, and movement direction.
You are given 0-indexed integer arrays positions, healths, and a string directions (directions[i] is either 'L' for left or 'R' for right). All integers in positions are unique.
All robots start moving on the line simultaneously at the same speed in their given directions. If two robots ever share the same position while moving, they will collide.
If two robots collide, the robot with lower health is removed from the line, and the health of the other robot decreases by one. The surviving robot continues in the same direction it was going. If both robots have the same health, they are both removed from the line.
Your task is to determine the health of the robots that survive the collisions, in the same order that the robots were given, i.e. final health of robot 1 (if survived), final health of robot 2 (if survived), and so on. If there are no survivors, return an empty array.
Return an array containing the health of the remaining robots (in the order they were given in the input), after no further collisions can occur.
Note: The positions may be unsorted.
Example 1:
Input: positions = [5,4,3,2,1], healths = [2,17,9,15,10], directions = "RRRRR" Output: [2,17,9,15,10] Explanation: No collision occurs in this example, since all robots are moving in the same direction. So, the health of the robots in order from the first robot is returned, [2, 17, 9, 15, 10].
Example 2:
Input: positions = [3,5,2,6], healths = [10,10,15,12], directions = "RLRL" Output: [14] Explanation: There are 2 collisions in this example. Firstly, robot 1 and robot 2 will collide, and since both have the same health, they will be removed from the line. Next, robot 3 and robot 4 will collide and since robot 4's health is smaller, it gets removed, and robot 3's health becomes 15 - 1 = 14. Only robot 3 remains, so we return [14].
Example 3:
Input: positions = [1,2,5,6], healths = [10,10,11,11], directions = "RLRL" Output: [] Explanation: Robot 1 and robot 2 will collide and since both have the same health, they are both removed. Robot 3 and 4 will collide and since both have the same health, they are both removed. So, we return an empty array, [].
Constraints:
1 <= positions.length == healths.length == directions.length == n <= 1051 <= positions[i], healths[i] <= 109directions[i] == 'L' or directions[i] == 'R'positions are distinctThis approach involves sorting the robots based on their positions so that we can simulate their movements along the line. We utilize a stack to keep track of the robots moving rightward. When a robot moving left encounters one or more robots on the stack, a collision is detected, and their healths are compared to determine the outcome.
The function first combines the input into a list of tuples and sorts it based on positions. For each robot, if it's moving right, its index is pushed onto a stack. When a left-moving robot encounters right-moving ones, collisions occur, and appropriate robots are removed based on their healths.
Java
Time Complexity: O(n log n) due to sorting the list of robots.
Space Complexity: O(n) because of the additional data structures used (stack and survivor list).
This approach simulates the collision process using a two-pointer technique. Each pointer represents a different direction of movement. By carefully checking each pair of robots, we can simulate their interactions and resolve collisions by adjusting health values and determining survivors.
This solution leverages a two-pointer-like mechanism where rightward-moving robots are managed with a stack. As each leftward-moving robot approaches, they are compared against the stack, with resulting health and survival adjustments.
C
Time Complexity: O(n log n) due to sorting the positions.
Space Complexity: O(n) due to extra data structures for handling collisions.
| Approach | Complexity |
|---|---|
| Sorting and Stack-based Collision Detection | Time Complexity: O(n log n) due to sorting the list of robots. |
| Two-Pointer Collision Simulation | Time Complexity: O(n log n) due to sorting the positions. |
Asteroid Collision - Stack - Leetcode 735 • NeetCode • 57,584 views views
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