solution and notes for problem 2751

Author: paxtonfitzpatrick

problems/2751/paxtonfitzpatrick.md

@@ -2,10 +2,94 @@
 
 ## Initial thoughts (stream-of-consciousness)
 
+- all robots are moving at the same speed, so any moving in same direction can't collide
+  - if all characters in `directions` are the same, can just return `healths`
+- also, any moving away from each other won't collide
+  - after filtering for same direction, can filter remaining opposite-direction movers for instances where one moving `R` has a larger `directions` value than another moving `L`.
+- this feels complex enough that I think it might be worth writing a `Robot` class to track attributes for each robot
+- order in which collisions happens will be important, because removed robot then won't collide with subsequent ones it would've otherwise encountered
+  - maybe this means I'll want to simulate this in terms "timesteps" so I can make sure to execute collision logic in the right order
+  - though simulation solutions are rarely the most optimal, and it seems like leetcode usually includes a test case that'd cause simulation-based solutions to time out
+  - I think the scenario where a simulation would be bad is if the colloding robots start very far from each other, causing me to have to simulate a bunch of noop timesteps (and check whether any robots are in the same location for each of them) before something actually happens. But I think I can avoid this by -- at each timestep where a collision happens, as well as the first -- finding the minimum distance between collision-eligible pair of robots and immediately advancing time by that many steps.
+- instead of simulating individual timesteps, could I figure this out from the initial positions and direction of the robots?
+  - I think so, if I start with the rightmost robot moving right and collide it with further-right robots that're moving left until it's dead... at which point I could move onto the next-rightmost right-moving robot and collide it with further-right left-moving robots
+  - Would this work with a scenario like example 3, where there are two indepenedent, simultaneous collisions?
+    - collide robot 3 with robot 4
+      - if it survives, we know it'll be in the returned array and we can move onto robot 1
+      - if it dies, we also move onto robot 1
+    - in either case, next move onto robot 1, collide it with robot 2
+      - if it survives
+        - if robot 4 survived the robot 3 vs 4 collision
+          - collide robot 1 with robot 4
+          - the survivor will be in the returned array
+        - if robot 3 survived the robot 3 vs 4 collision
+          - robot 1 will be in the returned array
+      - if it dies, robot 2 *and* robot 4 will be in the returned array
+  - I *think* this will work... I think I'll just need 2 separate lists(?) to keep track of right-moving and left-moving robots independently, so I can figure out which to collide with which
+    - also, both will need to be sorted so I can access the "next" robot from each list correctly
+    - I'd need to loop over robots to construct those two list anyway... if I pre-sort the initial list of robots moving in both directions, can I just do the collisions at the same time?
+    - this feels like it's trying to get me to use a stack... or two?
+  - let's say I get the list of robots in position order (I'll figure out how later). Then I could:
+    - create a stack for right-moving robots (and one for left-moving robots?)
+    - for each robot
+      - if it's moving right, push it onto a stack of right-moving robots
+        - this means the stack will also be in order of position, with the rightmost right-moving robot on top... yeah this definitely feels like how they're trying to get me to solve it... I think
+      - else (it's moving left), check whether there are any right-moving robots on the stack
+        - if there aren't, we know that robot is a survivor, and can move onto the next robot in the list
+        - else (there are) collide it with the right-moving robot on top of the stack
+          - if the left-moving robot survives, pop the right-moving robot off the stack (can discard because we know it didn't survive) and collide the left-moving robot with the new top-most right-moving robot, and so on
+            - (I think this means I'll actually need to use a `while` loop for checking whether there are robots in the stack instead of `if`/`else`)
+          - else (the right-moving robot survived) move onto the next robot in the sorted list and repeat
+          - ah, also need to account for equal health -- neither survives in that case
+    - any robots on the stack when we get to the end of the list are survivors
+    - also, seems like I *won't* need a separate stack for left-moving robots
+- how do i get the survivor list back into the right order?
+  - if I create a bunch of `Robot` objects, I could initialize them with an attr for their original index, and then sort the survivor list by `key=lambda robot: robot.<that_attr>`
+    - but the constraints say there can be up to $10^5$ robots, so creating that many objects -- even small, optimized ones with `namedtuple` or `dataclass` -- would take a ton of memory and also initialization time...
+    - what value do I keep track of in the stack of right-moving robots if I'm not using `Robot` objects though?
+  - could I create some sort of mapping between the robots' pre- and post-sorting indices?
+  - `np.argsort` would be nice here... alas
+    - could I create the equivalent output some other way?
+    - I could create a list of indices, and then sort that according to `positions` without *actually* sorting positions... then instead of sorting `positions` in the first place and mapping thos indices onto `healths`, `directions` , then trying to map backwards, I could use the sorted list of indices to get the next index from all 3 input lists for the current iteration.
+      - ooo important note with this though -- I wouldn't want to actually remove any defeated robots from any of the lists while iterating, because that would mess up the mapping. Also, once I determine the "current" robot is a survivor, I can't just append them to the list I'll eventually return because I'm iterating in sorted order and not input order.
+        - *But* all the input lists are still in their original order, including `healths`, which is the basis for the values I need to return... so I could just modify `healths` in-place and return that at the end.
+          - I'd have to filter 0's to remove defeated robots before returning it... which ultimately means another loop, so this might just be a wash. But it's the idea I have right now so I'll go with it...
+
 ## Refining the problem, round 2 thoughts
 
 ## Attempted solution(s)
+
 ```python
-class Solution:  # paste your code here!
-    ...
+class Solution:
+    def survivedRobotsHealths(self, positions: List[int], healths: List[int], directions: str) -> List[int]:
+        sorted_ixs = list(range(len(positions)))
+        sorted_ixs.sort(key=lambda i: positions[i])
+        right_movers_stack = []
+
+        for curr_ix in sorted_ixs:
+            if directions[curr_ix] == 'R':
+                right_movers_stack.append(curr_ix)
+            else:
+                while len(right_movers_stack) > 0:
+                    rightmost_right_mover_ix = right_movers_stack[-1]
+                    rightmost_right_mover_health = healths[rightmost_right_mover_ix]
+                    if rightmost_right_mover_health > healths[curr_ix]:
+                        healths[curr_ix] = 0
+                        healths[rightmost_right_mover_ix] -= 1
+                        break
+                    elif rightmost_right_mover_health == healths[curr_ix]:
+                        healths[curr_ix] = 0
+                        healths[rightmost_right_mover_ix] = 0
+                        right_movers_stack.pop()
+                        break
+                    else:
+                        right_movers_stack.pop()
+                        healths[rightmost_right_mover_ix] = 0
+                        healths[curr_ix] -= 1
+
+        return list(filter(None, healths))
 ```
+
+![](https://github.com/user-attachments/assets/8a60614f-f696-4922-8308-887508c6f70c)
+
+yay!