LeetCode 3367: Maximize Sum of Weights after Edge Removals Solution
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3367. Maximize Sum of Weights after Edge Removals
Problem Explanation
Explanation:
To maximize the sum of weights after edge removals, we need to ensure that each node has edges with at most k other nodes. This means we should remove edges strategically to meet this constraint while maximizing the sum of weights.
We can approach this problem using a depth-first search (DFS) on the tree. During the DFS traversal, we can keep track of the number of children each node has. If the number of children exceeds k, we remove the edge with the minimum weight connected to that node.
This way, we ensure that each node has edges with at most k other nodes while maximizing the sum of weights. We can implement this algorithm efficiently by sorting the edges by weight in descending order and then performing DFS.
Time Complexity: O(n log n) where n is the number of nodes in the tree. Sorting the edges takes O(n log n) time, and the DFS traversal takes O(n) time.
Space Complexity: O(n) for storing the tree structure and the stack space used for DFS.
:
Solution Code
class Solution {
public int maxWeight(int[][] edges, int k) {
Map<Integer, List<int[]>> graph = new HashMap<>();
for (int[] edge : edges) {
graph.computeIfAbsent(edge[0], key -> new ArrayList<>()).add(new int[]{edge[1], edge[2]});
graph.computeIfAbsent(edge[1], key -> new ArrayList<>()).add(new int[]{edge[0], edge[2]});
}
int[] res = new int[1];
dfs(graph, -1, 0, k, res);
return res[0];
}
private int dfs(Map<Integer, List<int[]>> graph, int parent, int node, int k, int[] res) {
List<Integer> children = new ArrayList<>();
for (int[] edge : graph.getOrDefault(node, Collections.emptyList())) {
if (edge[0] == parent) continue;
int childCount = dfs(graph, node, edge[0], k, res);
if (childCount > k) {
res[0] += edge[1];
} else {
children.add(childCount);
}
}
Collections.sort(children);
for (int i = 0; i < children.size() - k; i++) {
res[0] -= children.get(i);
}
return children.size() + 1;
}
}Try It Yourself
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