LeetCode #3367 — HARD

Maximize Sum of Weights after Edge Removals

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

There exists an undirected tree with n nodes numbered 0 to n - 1. You are given a 2D integer array edges of length n - 1, where edges[i] = [u_i, v_i, w_i] indicates that there is an edge between nodes u_i and v_i with weight w_i in the tree.

Your task is to remove zero or more edges such that:

Each node has an edge with at most k other nodes, where k is given.
The sum of the weights of the remaining edges is maximized.

Return the maximum possible sum of weights for the remaining edges after making the necessary removals.

Example 1:

Input: edges = [[0,1,4],[0,2,2],[2,3,12],[2,4,6]], k = 2

Output: 22

Explanation:

Node 2 has edges with 3 other nodes. We remove the edge [0, 2, 2], ensuring that no node has edges with more than k = 2 nodes.
The sum of weights is 22, and we can't achieve a greater sum. Thus, the answer is 22.

Example 2:

Input: edges = [[0,1,5],[1,2,10],[0,3,15],[3,4,20],[3,5,5],[0,6,10]], k = 3

Output: 65

Explanation:

Since no node has edges connecting it to more than k = 3 nodes, we don't remove any edges.
The sum of weights is 65. Thus, the answer is 65.

Constraints:

2 <= n <= 10⁵
1 <= k <= n - 1
edges.length == n - 1
edges[i].length == 3
0 <= edges[i][0] <= n - 1
0 <= edges[i][1] <= n - 1
1 <= edges[i][2] <= 10⁶
The input is generated such that edges form a valid tree.

Step 01

Brute Force Baseline

Problem summary: There exists an undirected tree with n nodes numbered 0 to n - 1. You are given a 2D integer array edges of length n - 1, where edges[i] = [ui, vi, wi] indicates that there is an edge between nodes ui and vi with weight wi in the tree. Your task is to remove zero or more edges such that: Each node has an edge with at most k other nodes, where k is given. The sum of the weights of the remaining edges is maximized. Return the maximum possible sum of weights for the remaining edges after making the necessary removals.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming · Tree

Example 1

[[0,1,4],[0,2,2],[2,3,12],[2,4,6]]
2

Example 2

[[0,1,5],[1,2,10],[0,3,15],[3,4,20],[3,5,5],[0,6,10]]
3

Core Insight

What unlocks the optimal approach

Can we use DFS based approach here?
For each edge, find two sums: one including the edge and one excluding it.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3367: Maximize Sum of Weights after Edge Removals
class Solution {
    private List<int[]>[] g;
    private int k;

    public long maximizeSumOfWeights(int[][] edges, int k) {
        this.k = k;
        int n = edges.length + 1;
        g = new List[n];
        Arrays.setAll(g, i -> new ArrayList<>());
        for (var e : edges) {
            int u = e[0], v = e[1], w = e[2];
            g[u].add(new int[] {v, w});
            g[v].add(new int[] {u, w});
        }
        var ans = dfs(0, -1);
        return Math.max(ans[0], ans[1]);
    }

    private long[] dfs(int u, int fa) {
        long s = 0;
        List<Long> t = new ArrayList<>();
        for (var e : g[u]) {
            int v = e[0], w = e[1];
            if (v == fa) {
                continue;
            }
            var res = dfs(v, u);
            s += res[0];
            long d = w + res[1] - res[0];
            if (d > 0) {
                t.add(d);
            }
        }
        t.sort(Comparator.reverseOrder());
        for (int i = 0; i < Math.min(t.size(), k - 1); ++i) {
            s += t.get(i);
        }
        return new long[] {s + (t.size() >= k ? t.get(k - 1) : 0), s};
    }
}

// Accepted solution for LeetCode #3367: Maximize Sum of Weights after Edge Removals
func maximizeSumOfWeights(edges [][]int, k int) int64 {
	n := len(edges) + 1
	g := make([][][]int, n)
	for _, e := range edges {
		u, v, w := e[0], e[1], e[2]
		g[u] = append(g[u], []int{v, w})
		g[v] = append(g[v], []int{u, w})
	}

	var dfs func(u, fa int) (int64, int64)
	dfs = func(u, fa int) (int64, int64) {
		var s int64
		var t []int64
		for _, e := range g[u] {
			v, w := e[0], e[1]
			if v == fa {
				continue
			}
			a, b := dfs(v, u)
			s += a
			d := int64(w) + b - a
			if d > 0 {
				t = append(t, d)
			}
		}
		sort.Slice(t, func(i, j int) bool {
			return t[i] > t[j]
		})
		for i := 0; i < min(len(t), k-1); i++ {
			s += t[i]
		}
		s2 := s
		if len(t) >= k {
			s += t[k-1]
		}
		return s, s2
	}

	x, y := dfs(0, -1)
	return max(x, y)
}

# Accepted solution for LeetCode #3367: Maximize Sum of Weights after Edge Removals
class Solution:
    def maximizeSumOfWeights(self, edges: List[List[int]], k: int) -> int:
        def dfs(u: int, fa: int) -> Tuple[int, int]:
            s = 0
            t = []
            for v, w in g[u]:
                if v == fa:
                    continue
                a, b = dfs(v, u)
                s += a
                if (d := (w + b - a)) > 0:
                    t.append(d)
            t.sort(reverse=True)
            return s + sum(t[:k]), s + sum(t[: k - 1])

        n = len(edges) + 1
        g: List[List[Tuple[int, int]]] = [[] for _ in range(n)]
        for u, v, w in edges:
            g[u].append((v, w))
            g[v].append((u, w))
        x, y = dfs(0, -1)
        return max(x, y)

// Accepted solution for LeetCode #3367: Maximize Sum of Weights after Edge Removals
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3367: Maximize Sum of Weights after Edge Removals
// class Solution {
//     private List<int[]>[] g;
//     private int k;
// 
//     public long maximizeSumOfWeights(int[][] edges, int k) {
//         this.k = k;
//         int n = edges.length + 1;
//         g = new List[n];
//         Arrays.setAll(g, i -> new ArrayList<>());
//         for (var e : edges) {
//             int u = e[0], v = e[1], w = e[2];
//             g[u].add(new int[] {v, w});
//             g[v].add(new int[] {u, w});
//         }
//         var ans = dfs(0, -1);
//         return Math.max(ans[0], ans[1]);
//     }
// 
//     private long[] dfs(int u, int fa) {
//         long s = 0;
//         List<Long> t = new ArrayList<>();
//         for (var e : g[u]) {
//             int v = e[0], w = e[1];
//             if (v == fa) {
//                 continue;
//             }
//             var res = dfs(v, u);
//             s += res[0];
//             long d = w + res[1] - res[0];
//             if (d > 0) {
//                 t.add(d);
//             }
//         }
//         t.sort(Comparator.reverseOrder());
//         for (int i = 0; i < Math.min(t.size(), k - 1); ++i) {
//             s += t.get(i);
//         }
//         return new long[] {s + (t.size() >= k ? t.get(k - 1) : 0), s};
//     }
// }

// Accepted solution for LeetCode #3367: Maximize Sum of Weights after Edge Removals
function maximizeSumOfWeights(edges: number[][], k: number): number {
    const n = edges.length + 1;
    const g: [number, number][][] = Array.from({ length: n }, () => []);
    for (const [u, v, w] of edges) {
        g[u].push([v, w]);
        g[v].push([u, w]);
    }
    const dfs = (u: number, fa: number): [number, number] => {
        let s = 0;
        const t: number[] = [];

        for (const [v, w] of g[u]) {
            if (v === fa) continue;

            const [a, b] = dfs(v, u);
            s += a;
            const d = w + b - a;
            if (d > 0) t.push(d);
        }

        t.sort((a, b) => b - a);
        for (let i = 0; i < Math.min(t.length, k - 1); i++) {
            s += t[i];
        }
        const s2 = s;
        if (t.length >= k) {
            s += t[k - 1];
        }

        return [s, s2];
    };

    const [x, y] = dfs(0, -1);
    return Math.max(x, y);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.