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

Author: jeremymanning

problems/3577/gpt5-mini.md

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+# [Problem 3577: Count the Number of Computer Unlocking Permutations](https://leetcode.com/problems/count-the-number-of-computer-unlocking-permutations/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+We are given an array complexity of length n and asked how many permutations of [0..n-1] represent valid unlocking orders starting from computer 0 being already decrypted (the root).  
+The unlocking rule for computer i is: at the moment you try to unlock i, there must already be some unlocked computer j with j < i and complexity[j] < complexity[i].
+
+I need to carefully interpret the rule: j is a label (an integer less than i), and j must be among already unlocked computers. Computer 0 is initially unlocked, so for any i>0, j = 0 is a candidate provided complexity[0] < complexity[i]. If complexity[0] is strictly smaller than complexity[i], then 0 can always unlock i at any time (because 0 < i always and 0 is unlocked from the start). If complexity[0] is not strictly smaller than some i (equal or larger), then that i cannot be unlocked by 0; it would require some other j < i with smaller complexity, but if no such j exists among labels < i then i can never be unlocked.
+
+This suggests a simple necessary condition: for every i > 0 there must exist some k < i with complexity[k] < complexity[i]. In particular for i = 1 this requires complexity[0] < complexity[1]. For i = 2 this requires min(complexity[0], complexity[1]) < complexity[2], etc. Observing the chain, if complexity[0] is not strictly the global minimum (i.e., there's some value <= complexity[0] elsewhere), then there will be some index that cannot be unlocked. So I suspect complexity[0] must be the unique minimum in the whole array. If it is the unique minimum, then 0 can unlock every other node immediately, so after placing 0 first all remaining nodes are available in any order, giving (n-1)! permutations (with 0 first).
+
+I should double-check examples and edge cases (duplicates of minimum, etc.).
+
+## Refining the problem, round 2 thoughts
+- If complexity[0] is not strictly smaller than all other complexities, there exists some index i>0 such that no j<i has complexity[j] < complexity[i] (in particular if complexity[i] <= complexity[0] and all earlier ones >= complexity[i]) — that i can never be unlocked. So answer 0 in that case.
+- If complexity[0] is the unique global minimum, then for every i>0 we have complexity[0] < complexity[i] and 0 < i, so 0 can unlock any i at any time. Once 0 is placed (the initial unlocked computer, logically first in unlocking order), all remaining n-1 computers are simultaneously available and can be unlocked in any permutation: (n-1)! ways.
+- Hence the solution becomes: check if complexity[0] is the unique smallest element in the array. If not, return 0. Otherwise return factorial(n-1) modulo 1e9+7.
+- Time complexity: O(n) to find min and count and O(n) to compute factorial up to n-1 (or O(1) with precomputation if cached). Space O(1).
+
+This reasoning matches the provided examples: [1,2,3] -> complexity[0] unique min -> (3-1)! = 2. [3,3,3,4,4,4] -> complexity[0] not unique min (ties) -> 0.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def countPermutations(self, complexity: List[int]) -> int:
+        MOD = 10**9 + 7
+        n = len(complexity)
+        # Check if complexity[0] is the unique global minimum
+        min_val = min(complexity)
+        if complexity[0] != min_val:
+            return 0
+        # If the minimum occurs more than once, 0 cannot be unlocked by 0 (strict < required)
+        if complexity.count(min_val) != 1:
+            return 0
+        # Otherwise, 0 can unlock any other computer, so the remaining (n-1) can be ordered arbitrarily
+        res = 1
+        for i in range(2, n):  # multiply 2 * 3 * ... * (n-1)
+            res = (res * i) % MOD
+        return res
+```
+- Notes on the solution:
+  - We check that complexity[0] equals the minimum and that the minimum is unique. If either condition fails, return 0.
+  - If the condition holds, the number of valid permutations is (n-1)! modulo 1e9+7 because after the (implicit) first unlocked computer 0, all other n-1 computers can be unlocked in any order (0 can directly unlock each of them).
+  - Time complexity: O(n) for finding min, counting occurrences, and computing factorial up to n-1 (overall O(n)). Space complexity: O(1) extra space.