[최원준] 2026/10/04

leetcode3/최원준/2322. Minimum Score After Removals on a Tree.py

@@ -0,0 +1,74 @@
+class Solution:
+    def minimumScore(self, nums: list[int], edges: list[list[int]]) -> int:
+        n = len(nums)
+
+        graph = [[] for _ in range(n)]
+
+        for a, b in edges:
+            graph[a].append(b)
+            graph[b].append(a)
+
+        sub_xor = [0] * n
+        tin = [0] * n
+        tout = [0] * n
+
+        time = 0
+
+        def dfs(node, parent):
+            nonlocal time
+
+            tin[node] = time
+            time += 1
+
+            xor_value = nums[node]
+
+            for nxt in graph[node]:
+                if nxt == parent:
+                    continue
+
+                dfs(nxt, node)
+                xor_value ^= sub_xor[nxt]
+
+            sub_xor[node] = xor_value
+
+            tout[node] = time
+            time += 1
+
+        dfs(0, -1)
+
+        total = sub_xor[0]
+
+        def is_ancestor(a, b):
+            return tin[a] <= tin[b] and tout[b] <= tout[a]
+
+        ans = float("inf")
+
+        # node i는
+        # parent(i) - i 간선을 자른다는 의미
+        #
+        # root(0)는 부모 간선이 없으므로 제외
+        for a in range(1, n):
+            for b in range(a + 1, n):
+
+                if is_ancestor(a, b):
+                    # a 안에 b가 존재
+                    x1 = sub_xor[b]
+                    x2 = sub_xor[a] ^ sub_xor[b]
+                    x3 = total ^ sub_xor[a]
+
+                elif is_ancestor(b, a):
+                    # b 안에 a가 존재
+                    x1 = sub_xor[a]
+                    x2 = sub_xor[b] ^ sub_xor[a]
+                    x3 = total ^ sub_xor[b]
+
+                else:
+                    # 서로 독립된 subtree
+                    x1 = sub_xor[a]
+                    x2 = sub_xor[b]
+                    x3 = total ^ sub_xor[a] ^ sub_xor[b]
+
+                score = max(x1, x2, x3) - min(x1, x2, x3)
+                ans = min(ans, score)
+
+        return ans