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

Author: jeremymanning

problems/1437/gpt5-mini.md

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+# [Problem 1437: Check If All 1's Are at Least Length K Places Away](https://leetcode.com/problems/check-if-all-1s-are-at-least-length-k-places-away/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+I need to verify that every pair of 1s in the array has at least k zeros between them. The array length up to 1e5 so an O(n) scan is fine. The simplest idea: walk the array, remember the index of the previous 1; when I encounter a new 1, compute the gap between indices and check whether there are at least k zeros between them. If any gap is too small, return False. If I finish the scan, return True.
+
+Edge quick thoughts: if k == 0 it's always true because no separation required. For the first encountered 1 there is no previous to compare with. Use -inf or None to indicate no previous 1. Be careful with off-by-one: if previous 1 at i_prev and current at i, number of places between = i - i_prev - 1, so require i - i_prev - 1 >= k -> equivalently i - i_prev > k.
+
+## Refining the problem, round 2 thoughts
+Implementation details:
+- Iterate with enumerate to get indices.
+- Keep prev index initialized to None.
+- On nums[i] == 1:
+  - If prev is not None, check if i - prev - 1 < k -> return False.
+  - Update prev = i.
+- Complexity: single pass, O(n) time, O(1) extra space.
+Alternative approaches: you could count consecutive zeros after each 1, or use a queue to store last 1s, but those are no simpler or not more efficient in this problem.
+
+Edge cases:
+- All zeros -> True.
+- Single 1 -> True.
+- k = 0 -> True.
+- Adjacent 1s -> must be k = 0 to be valid.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def kLengthApart(self, nums: List[int], k: int) -> bool:
+        prev = None  # index of previous 1
+        for i, val in enumerate(nums):
+            if val == 1:
+                if prev is not None:
+                    # number of places between prev and i is (i - prev - 1)
+                    if i - prev - 1 < k:
+                        return False
+                prev = i
+        return True
+```
+- Notes:
+  - Approach: single pass keeping the index of the last seen 1 and checking the number of zeros between consecutive 1s.
+  - Time complexity: O(n), where n = len(nums), because we examine each element once.
+  - Space complexity: O(1) extra space (only an index variable).
+  - Implementation detail: the check uses i - prev - 1 < k (equivalently i - prev <= k) to detect a violation.