Minimum Wizard Referral Cost
Reported by candidates from Airbnb's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The whole problem hinges on a priority queue. Airbnb reported this one in March 2019, and if your OA invite is a day or two out, know it now: it's a weighted directed graph dressed up as wizards and referrals. Edges out of wizard 0 cost nothing, every other edge costs (j - i) squared, and you want the cheapest way to reach wizard n - 1. Cycles are allowed, so a naive recursion will hang. If you blank on the setup during the live OA, StealthCoder runs invisibly as a safety net and hands you the approach.
The problem
There are n wizards ranked from 0 through n - 1. The string referrals[i] contains the space-separated ranks of the wizards whom wizard i can introduce. An empty string means that wizard i has no referrals. You are wizard 0. You already know every wizard listed by wizard 0, so following an edge from wizard 0 costs 0. For every other directed referral from wizard i to wizard j, the introduction costs (j - i)2. Return the minimum total cost needed to meet the highest-ranked wizard, n - 1. Return -1 if that wizard cannot be reached. The referral graph may contain cycles. Function minimumWizardReferralCost(referrals: String[]) → int Examples Example 1 referrals = ["1 2","3","3",""] return = 1 Wizards 1 and 2 are both known for free. Asking wizard 2 to introduce wizard 3 costs (3 - 2)^2 = 1, which is cheaper than the cost 4 route through wizard 1. Example 2 referrals = ["3","","",""] return = 0 Wizard 0 already knows the highest-ranked wizard directly, so no paid introduction is needed. Example 3 referrals = ["1","0",""] return = -1 The cycle between wizards 0 and 1 never reaches wizard 2. Constraints 2 <= referrals.length <= 500. Each non-empty row contains distinct integer ranks separated by one space. Every listed rank is in [0, referrals.length - 1]. The graph may contain directed cycles.
Reported by candidates. Source: FastPrep
Pattern and pitfall
This is single-source shortest path with non-negative weights, so Dijkstra with a min-heap is the tool. Parse each referrals[i] by splitting on spaces, and skip empty strings so you don't crash on int(''). Edges leaving node 0 get weight 0. Every other edge gets (j - i)^2. Push (0, 0) into the heap, pop the cheapest, skip stale entries, relax neighbors, and return the distance to n - 1, or -1 if it's never reached. The common pitfall is applying the zero cost to edges that merely lead to wizard 0's neighbors instead of only edges leaving wizard 0. Example 1 shows this: going 2 to 3 costs 1, not 0. Another trap is the cycle in Example 3, which a visited check or distance comparison handles. With n up to 500, performance is fine. StealthCoder is your hedge if the heap code slips on the day.
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Minimum Wizard Referral Cost FAQ
What's the trick in Minimum Wizard Referral Cost?+
Treat it as a weighted directed graph and run Dijkstra. Edges out of wizard 0 cost 0, all other edges cost (j - i) squared. A min-heap gives you the cheapest path to wizard n - 1, and the same pass handles cycles safely.
Why can't I use plain BFS here?+
BFS counts edges, not cost. In Example 1 the route through wizard 2 costs 1 while the route through wizard 1 costs 4, though both have the same hop count. Weights differ per edge, so you need Dijkstra, not an unweighted traversal.
How do I handle empty referral strings?+
Splitting an empty string on a space gives a list with one empty item in some languages, which breaks integer parsing. Check that the string is non-empty, or filter out blank tokens, before converting to ints and adding edges.
When do I return -1?+
Return -1 if the heap empties and wizard n - 1 never got a finite distance. Example 3 shows it: wizards 0 and 1 point at each other, and wizard 2 is unreachable. Keep distances initialized to infinity and check at the end.
How should I prepare in 48 hours for this kind of OA?+
Write Dijkstra from memory twice with a heap, including the stale-entry skip. Then practice parsing odd input formats like space-separated strings inside an array. Those two things cover almost everything this Airbnb problem, reported March 2019, tests.