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Author: hogan-tech

2872-maximum-number-of-k-divisible-components/2872-maximum-number-of-k-divisible-components.py

@@ -0,0 +1,39 @@
+# time complexity: O(n+m)
+# space complexity: O(n+m)
+from typing import List
+
+
+class Solution:
+    def maxKDivisibleComponents(self, n: int, edges: List[List[int]], values: List[int], k: int) -> int:
+        adjList = [[] for _ in range(n)]
+        for node1, node2 in edges:
+            adjList[node1].append(node2)
+            adjList[node2].append(node1)
+
+        componentCount = [0]
+        self.dfs(0, -1, adjList, values, k, componentCount)
+
+        return componentCount[0]
+
+    def dfs(self, currNode: int, parentNode: int, adjList: List[List[int]],
+            nodeValues: List[int], k: int, componentCount: List[int]) -> int:
+
+        total = 0
+        for neighborNode in adjList[currNode]:
+            if neighborNode != parentNode:
+                total += self.dfs(neighborNode, currNode,
+                                  adjList, nodeValues, k, componentCount)
+                total %= k
+        total += nodeValues[currNode]
+        total %= k
+        if total == 0:
+            componentCount[0] += 1
+
+        return total
+
+
+n = 5
+edges = [[0, 2], [1, 2], [1, 3], [2, 4]]
+values = [1, 8, 1, 4, 4]
+k = 6
+print(Solution().maxKDivisibleComponents(n, edges, values, k))