0828MediumArrayGreedyGraph TheorySortingHeap (Priority Queue)

Best Hub and Spokes Yield

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Trains the technique from

LeetCode 2497Maximum Star Sum of a Graph

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A survey network has n sites numbered 0 through n - 1. Site i reports a yield of vals[i], which may be negative. edges[j] = [a, b] is a two-way cable between site a and site b.

A cluster is one site chosen as the hub, together with any set of at most k sites that share a cable with it. The hub is always part of its own cluster, and a cluster may consist of the hub alone. The yield of a cluster is the sum of the yields of its sites.

Return the largest yield any cluster can have.

Examples

Example 1

Input: vals = [2, -1, 5, -3], edges = [[0, 1], [0, 2], [1, 3]], k = 2
Output: 7

Taking site 0 as the hub and site 2 as its only spoke gives 2 + 5 = 7. Sites 0 and 2 do share a cable, and one spoke is within the allowance of two.

Example 2

Input: vals = [-5, -6, -7], edges = [[0, 1], [1, 2]], k = 2
Output: -5

Every yield is negative. A cluster of site 0 alone has yield -5, and no cluster does better.

Example 3

Input: vals = [10, 0, 0], edges = [[0, 1], [0, 2]], k = 2
Output: 10

Site 0 as the hub gives 10, and adding either zero-yield spoke leaves the total at 10, so 10 is the largest yield.

Constraints

1 <= vals.length <= 10^5
-10^4 <= vals[i] <= 10^4
0 <= edges.length <= 10^5
edges[j].length == 2
0 <= edges[j][0] <= 99999
0 <= edges[j][1] <= 99999
A cable never joins a site to itself, and every site index is below vals.length
0 <= k <= 99999

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def best_hub_yield(vals: list[int], edges: list[list[int]], k: int) -> int:

Java

public int bestHubYield(int[] vals, int[][] edges, int k)

The intended approach

Time O(n + E + sum over hubs of d log d) for degrees d, which is O(E log E) overall.Space O(n + E) for the neighbour yield lists.

The hub fixes which sites are available, and the spokes never interact, so each hub is settled on its own. Collect the yields of a hub's cabled neighbours, sort them descending, and take them while the count is below the allowance and the yield is still positive, since taking a zero or negative yield can never improve a total and the allowance is an upper bound rather than a quota. The best total over all hubs is the answer.

Pythonentry point: best_hub_yield

def best_hub_yield(vals, edges, k):
    n = len(vals)
    around = [[] for _ in range(n)]
    for a, b in edges:
        around[a].append(vals[b])
        around[b].append(vals[a])
    best = None
    for hub in range(n):
        total = vals[hub]
        picks = sorted(around[hub], reverse=True)
        taken = 0
        for yield_ in picks:
            if taken == k or yield_ <= 0:
                break
            total += yield_
            taken += 1
        if best is None or total > best:
            best = total
    return best

The obvious version

For every hub, enumerate every set of at most k of its cabled neighbours and total the yields, keeping the largest. Exponential in the degree, but it makes no assumption about which spokes are worth taking.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_best_hub_yield(vals, edges, k):
    import itertools
    n = len(vals)
    around = [[] for _ in range(n)]
    for a, b in edges:
        around[a].append(b)
        around[b].append(a)
    best = None
    for hub in range(n):
        mates = around[hub]
        for size in range(0, min(k, len(mates)) + 1):
            for combo in itertools.combinations(mates, size):
                total = vals[hub] + sum(vals[m] for m in combo)
                if best is None or total > best:
                    best = total
    return best

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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