How should I practice First Eligible Partner in a Social Graph?

Start by reading the full statement, solve the visible examples, then compare the pattern against related Bloomberg LP phone screen problems in FastPrep.

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FastPrep tags this as a Bloomberg LP phone screen coding practice problem when the problem catalog has that company and stage context.

First Eligible Partner in a Social Graph

Bloomberg LP● MediumNEW GRADPHONE SCREEN

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Problem statement

You are given an undirected social graph with people numbered from 0 through n - 1. Each row [u, v] in edges is one social link. The array eligible has length n, where eligible[i] == 1 means person i satisfies the partner condition.

Starting from start, return the eligible person reachable with the fewest social hops. Do not consider start as their own partner. If several eligible people are equally close, return the smaller person ID. Return -1 when no eligible partner is reachable.

Function

findFirstEligiblePartner(n: int, edges: int[][], start: int, eligible: int[]) → int

Examples

Example 1

n = 6edges = [[0,1],[0,2],[1,3],[2,4],[3,5]]start = 0eligible = [0,0,0,1,1,1]return = 3

People 3 and 4 are both two hops away. The smaller ID, 3, wins the tie.

Example 2

n = 4edges = [[0,1]]start = 0eligible = [1,0,1,0]return = -1

The starting person is excluded, and the only other eligible person is disconnected.

Example 3

n = 5edges = [[0,4],[0,2],[4,3]]start = 0eligible = [0,0,1,0,1]return = 2

People 2 and 4 are each one hop away, so the smaller ID is returned.

Constraints

1 <= n <= 100000.
0 <= edges.length <= 200000.
Every edge contains two different valid person IDs; edges are unique and represent undirected links.
eligible.length == n and every value is 0 or 1.
0 <= start < n.