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

Author: jeremymanning

problems/2872/gpt5-mini.md

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+# [Problem 2872: Maximum Number of K-Divisible Components](https://leetcode.com/problems/maximum-number-of-k-divisible-components/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+We have a tree and node values; we can remove some edges so that every resulting connected component has sum of node values divisible by k. The total sum is divisible by k (given), so at least the whole tree is valid (no cuts). Cutting an edge splits a subtree from the rest; the sum of that subtree must be divisible by k. That suggests we should look at subtree sums modulo k.
+
+If we root the tree somewhere (say node 0), for any non-root node u, if the sum of the subtree rooted at u is divisible by k, then we may cut the edge between u and its parent and get an extra valid component. Cutting such an edge doesn't break the property for other disjoint subtrees. Intuitively we can cut every edge whose child-subtree sum % k == 0 independently. So count how many non-root nodes have subtree sum % k == 0; answer is that count + 1 (components = cuts + 1). DFS to compute subtree sum modulo k seems natural.
+
+## Refining the problem, round 2 thoughts
+- Implementation detail: sums can be large but we only need them modulo k, so propagate s = (s + child_sum) % k.
+- When a subtree sum % k == 0 for node u (u != root), we can increment cut count and return 0 to parent (simulate cutting so parent doesn't include the subtree sum).
+- Edge cases:
+  - n == 1 -> no edges, answer should be 1.
+  - k == 1 -> every value % 1 == 0, every subtree sum % 1 == 0 -> we can cut all n-1 edges -> answer = n.
+- Complexity: one DFS over n nodes, O(n) time and O(n) space for adjacency + recursion stack. For Python recursion, increase recursion limit (n up to 3e4).
+- Correctness: cutting every edge where the child-subtree sum is 0 (mod k) is never harmful — those subtrees are independent of the rest and produce valid components. Since total sum divisible by k, root ends up with 0 mod k but we don't attempt to cut above root.
+
+## Attempted solution(s)
+```python
+import sys
+from typing import List
+
+class Solution:
+    def maxKDivisibleComponents(self, n: int, edges: List[List[int]], values: List[int], k: int) -> int:
+        # Build adjacency list
+        adj = [[] for _ in range(n)]
+        for u, v in edges:
+            adj[u].append(v)
+            adj[v].append(u)
+
+        sys.setrecursionlimit(1000000)
+        cuts = 0
+
+        def dfs(u: int, parent: int) -> int:
+            nonlocal cuts
+            # subtree sum modulo k
+            s = values[u] % k
+            for v in adj[u]:
+                if v == parent:
+                    continue
+                child_mod = dfs(v, u)
+                s = (s + child_mod) % k
+            # If this subtree (excluding parent) sums to 0 mod k and it's not the root,
+            # we can cut the edge to parent and return 0 upwards.
+            if parent != -1 and s == 0:
+                cuts += 1
+                return 0
+            return s
+
+        dfs(0, -1)
+        # components = cuts + 1 (initial component + each cut increases count by 1)
+        return cuts + 1
+
+# Example usage:
+# sol = Solution()
+# print(sol.maxKDivisibleComponents(5, [[0,2],[1,2],[1,3],[2,4]], [1,8,1,4,4], 6))  # -> 2
+```
+
+- Notes about the solution:
+  - Approach: Root the tree (node 0), DFS to compute subtree sums modulo k. Whenever a non-root subtree sums to 0 mod k, increment cut count and return 0 so parent doesn't include it.
+  - Time complexity: O(n) — each node/edge visited once.
+  - Space complexity: O(n) for adjacency list + O(n) recursion stack in worst case.
+  - Implementation details: use modulo arithmetic to avoid large integers; increase recursion limit for deep trees.