Auto-solve daily LeetCode problem using GPT-5-mini

Author: jeremymanning

problems/2069/gpt5-mini.md

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+# [Problem 2069: Walking Robot Simulation II](https://leetcode.com/problems/walking-robot-simulation-ii/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+The robot moves along the boundary of a width x height grid. Every time it would step out of bounds it turns 90° counterclockwise and keeps trying until it completes that step. Observing the motion, the robot always stays on the perimeter (border) cells and traverses them in a fixed cyclical order: along the bottom edge from left to right, then up the right edge, then along the top edge right-to-left, then down the left edge, and repeat.
+
+So we can precompute the list of perimeter coordinates in that traversal order and treat the robot as moving along indices of that list. A step(num) is simply advancing the current index by num (mod perimeter length). For getDir(), direction depends on which segment of the perimeter the current index lies on; there's a special-case: when the robot is at the starting position (index 0) and it has completed at least one step in the past (i.e., total steps taken > 0), it should be facing "South" (this matches the turn pattern after a full loop).
+
+This yields O(width+height) setup and O(1) per operation.
+
+## Refining the problem, round 2 thoughts
+Edge cases:
+- width, height >= 2 so perimeter length is at least 4.
+- step(num) can be large, so use modulo arithmetic to update index.
+- Need to know whether the robot has ever moved a positive number of steps to decide the special direction at index 0. Keep a total_steps counter (unbounded integer in Python is fine).
+- Compute the boundary list carefully to avoid duplicating corner cells.
+
+Time complexity:
+- Initialization: O(width + height)
+- step / getPos / getDir: O(1)
+
+Space complexity:
+- O(perimeter) = O(width + height)
+
+## Attempted solution(s)
+```python
+class Robot:
+    def __init__(self, width: int, height: int):
+        # Build the perimeter path in the order the robot will traverse:
+        # bottom row left->right, right column bottom->top,
+        # top row right->left, left column top->bottom (excluding corners duplicates)
+        self.w = width
+        self.h = height
+        path = []
+        # bottom edge (y = 0), x = 0..w-1
+        for x in range(0, width):
+            path.append((x, 0))
+        # right edge (x = w-1), y = 1..h-1
+        for y in range(1, height):
+            path.append((width - 1, y))
+        # top edge (y = h-1), x = w-2..0
+        for x in range(width - 2, -1, -1):
+            path.append((x, height - 1))
+        # left edge (x = 0), y = h-2..1
+        for y in range(height - 2, 0, -1):
+            path.append((0, y))
+
+        self.path = path
+        self.L = len(path)  # perimeter length, equals 2*(w + h) - 4
+        self.idx = 0        # current index in path
+        self.total_steps = 0  # total steps ever taken (to detect full-loop special case)
+
+    def step(self, num: int) -> None:
+        if self.L == 0:
+            return
+        self.total_steps += num
+        self.idx = (self.idx + num) % self.L
+
+    def getPos(self) -> [int, int]:
+        x, y = self.path[self.idx]
+        return [x, y]
+
+    def getDir(self) -> str:
+        # Special case: if at starting cell after having moved >0 steps, robot faces South
+        if self.idx == 0 and self.total_steps > 0:
+            return "South"
+
+        # Determine which edge segment the current index is on
+        # indices:
+        # 0 .. (w-1) - 1        -> bottom edge -> East
+        # (w-1) .. (w-1 + h-1) - 1 -> right edge -> North
+        # ... next (w-1) -> top edge -> West
+        # remaining -> left edge -> South
+        seg1 = self.w - 1
+        seg2 = seg1 + (self.h - 1)
+        seg3 = seg2 + (self.w - 1)
+        # seg4 is implicit remainder
+
+        if self.idx < seg1:
+            return "East"
+        elif self.idx < seg2:
+            return "North"
+        elif self.idx < seg3:
+            return "West"
+        else:
+            return "South"
+```
+- Notes:
+  - We precompute the perimeter path in traversal order; moving is an index shift modulo path length.
+  - We maintain total_steps to distinguish the initial facing ("East" at start index 0) from the facing after a full loop (index 0 but should be "South").
+  - Time: O(width + height) init, O(1) per method call.
+  - Space: O(width + height) for the path list.