LeetCode #2497 — MEDIUM

Maximum Star Sum of a Graph

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

There is an undirected graph consisting of n nodes numbered from 0 to n - 1. You are given a 0-indexed integer array vals of length n where vals[i] denotes the value of the i^th node.

You are also given a 2D integer array edges where edges[i] = [a_i, b_i] denotes that there exists an undirected edge connecting nodes a_i and b_i.

A star graph is a subgraph of the given graph having a center node containing 0 or more neighbors. In other words, it is a subset of edges of the given graph such that there exists a common node for all edges.

The image below shows star graphs with 3 and 4 neighbors respectively, centered at the blue node.

The star sum is the sum of the values of all the nodes present in the star graph.

Given an integer k, return the maximum star sum of a star graph containing at most k edges.

Example 1:

Input: vals = [1,2,3,4,10,-10,-20], edges = [[0,1],[1,2],[1,3],[3,4],[3,5],[3,6]], k = 2
Output: 16
Explanation: The above diagram represents the input graph.
The star graph with the maximum star sum is denoted by blue. It is centered at 3 and includes its neighbors 1 and 4.
It can be shown it is not possible to get a star graph with a sum greater than 16.

Example 2:

Input: vals = [-5], edges = [], k = 0
Output: -5
Explanation: There is only one possible star graph, which is node 0 itself.
Hence, we return -5.

Constraints:

n == vals.length
1 <= n <= 10⁵
-10⁴ <= vals[i] <= 10⁴
0 <= edges.length <= min(n * (n - 1) / 2, 10⁵)
edges[i].length == 2
0 <= a_i, b_i <= n - 1
a_i != b_i
0 <= k <= n - 1

Step 01

Brute Force Baseline

Problem summary: There is an undirected graph consisting of n nodes numbered from 0 to n - 1. You are given a 0-indexed integer array vals of length n where vals[i] denotes the value of the ith node. You are also given a 2D integer array edges where edges[i] = [ai, bi] denotes that there exists an undirected edge connecting nodes ai and bi. A star graph is a subgraph of the given graph having a center node containing 0 or more neighbors. In other words, it is a subset of edges of the given graph such that there exists a common node for all edges. The image below shows star graphs with 3 and 4 neighbors respectively, centered at the blue node. The star sum is the sum of the values of all the nodes present in the star graph. Given an integer k, return the maximum star sum of a star graph containing at most k edges.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

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

Example 2

[-5]
[]
0

Core Insight

What unlocks the optimal approach

A star graph doesn’t necessarily include all of its neighbors.
For each node, sort its neighbors in descending order and take k max valued neighbors.

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

Upper-end input sizes

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 #2497: Maximum Star Sum of a Graph
class Solution {
    public int maxStarSum(int[] vals, int[][] edges, int k) {
        int n = vals.length;
        List<Integer>[] g = new List[n];
        Arrays.setAll(g, key -> new ArrayList<>());
        for (var e : edges) {
            int a = e[0], b = e[1];
            if (vals[b] > 0) {
                g[a].add(vals[b]);
            }
            if (vals[a] > 0) {
                g[b].add(vals[a]);
            }
        }
        for (var e : g) {
            Collections.sort(e, (a, b) -> b - a);
        }
        int ans = Integer.MIN_VALUE;
        for (int i = 0; i < n; ++i) {
            int v = vals[i];
            for (int j = 0; j < Math.min(g[i].size(), k); ++j) {
                v += g[i].get(j);
            }
            ans = Math.max(ans, v);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2497: Maximum Star Sum of a Graph
func maxStarSum(vals []int, edges [][]int, k int) (ans int) {
	n := len(vals)
	g := make([][]int, n)
	for _, e := range edges {
		a, b := e[0], e[1]
		if vals[b] > 0 {
			g[a] = append(g[a], vals[b])
		}
		if vals[a] > 0 {
			g[b] = append(g[b], vals[a])
		}
	}
	for _, e := range g {
		sort.Sort(sort.Reverse(sort.IntSlice(e)))
	}
	ans = math.MinInt32
	for i, v := range vals {
		for j := 0; j < min(len(g[i]), k); j++ {
			v += g[i][j]
		}
		ans = max(ans, v)
	}
	return
}

# Accepted solution for LeetCode #2497: Maximum Star Sum of a Graph
class Solution:
    def maxStarSum(self, vals: List[int], edges: List[List[int]], k: int) -> int:
        g = defaultdict(list)
        for a, b in edges:
            if vals[b] > 0:
                g[a].append(vals[b])
            if vals[a] > 0:
                g[b].append(vals[a])
        for bs in g.values():
            bs.sort(reverse=True)
        return max(v + sum(g[i][:k]) for i, v in enumerate(vals))

// Accepted solution for LeetCode #2497: Maximum Star Sum of a Graph
// 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 #2497: Maximum Star Sum of a Graph
// class Solution {
//     public int maxStarSum(int[] vals, int[][] edges, int k) {
//         int n = vals.length;
//         List<Integer>[] g = new List[n];
//         Arrays.setAll(g, key -> new ArrayList<>());
//         for (var e : edges) {
//             int a = e[0], b = e[1];
//             if (vals[b] > 0) {
//                 g[a].add(vals[b]);
//             }
//             if (vals[a] > 0) {
//                 g[b].add(vals[a]);
//             }
//         }
//         for (var e : g) {
//             Collections.sort(e, (a, b) -> b - a);
//         }
//         int ans = Integer.MIN_VALUE;
//         for (int i = 0; i < n; ++i) {
//             int v = vals[i];
//             for (int j = 0; j < Math.min(g[i].size(), k); ++j) {
//                 v += g[i].get(j);
//             }
//             ans = Math.max(ans, v);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2497: Maximum Star Sum of a Graph
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2497: Maximum Star Sum of a Graph
// class Solution {
//     public int maxStarSum(int[] vals, int[][] edges, int k) {
//         int n = vals.length;
//         List<Integer>[] g = new List[n];
//         Arrays.setAll(g, key -> new ArrayList<>());
//         for (var e : edges) {
//             int a = e[0], b = e[1];
//             if (vals[b] > 0) {
//                 g[a].add(vals[b]);
//             }
//             if (vals[a] > 0) {
//                 g[b].add(vals[a]);
//             }
//         }
//         for (var e : g) {
//             Collections.sort(e, (a, b) -> b - a);
//         }
//         int ans = Integer.MIN_VALUE;
//         for (int i = 0; i < n; ++i) {
//             int v = vals[i];
//             for (int j = 0; j < Math.min(g[i].size(), k); ++j) {
//                 v += g[i].get(j);
//             }
//             ans = Math.max(ans, v);
//         }
//         return ans;
//     }
// }

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 log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.