Vectorized K-Means Assignments
Reported by candidates from SambaNova Systems's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
SambaNova Systems reported this one in April 2024, and the trap is hiding in the tie-break and empty-cluster rules, not the math. Vectorized K-Means Assignments looks like a textbook loop, so people write it from memory and fail on the edge cases. It's a simulation problem: initialize centroids from the first k points, assign, recompute, repeat until nothing moves. If you blank on the details during the live OA, StealthCoder runs invisibly as a safety net and reads the problem for you. Better to know the traps going in, though. There are three, and they're all in the spec.
The problem
Implement deterministic K-Means clustering for integer points. Use the first k points as the initial centroids, in that order. Repeat these steps until no point changes its assigned cluster: Assign every point to the centroid with the smallest squared Euclidean distance. Break a tie by choosing the smaller cluster index. Replace each nonempty cluster's centroid with the coordinate-wise arithmetic mean of its assigned points. If a cluster is empty, keep its previous centroid. Return an integer array whose value at index i is the zero-based cluster index assigned to points[i]. Implement the arithmetic and iteration directly; do not call a clustering library. Function clusterPoints(points: int[][], k: int) → int[] Examples Example 1 points = [[0,0],[10,10],[1,0],[9,10]] k = 2 return = [0,1,0,1] The first two points initialize the two centroids. The two points near the origin join cluster 0, and the two points near [10,10] join cluster 1. Recomputed centroids preserve those assignments. Example 2 points = [[0],[4],[2]] k = 2 return = [0,1,0] The point [2] is initially equally distant from centroids 0 and 4, so the smaller cluster index wins the tie. Example 3 points = [[-5,2],[7,-1],[3,4]] k = 3 return = [0,1,2] When every point initializes its own cluster, each point remains at distance zero from that centroid. Constraints 1 <= points.length <= 200 1 <= points[i].length <= 10 All points have the same dimension. 1 <= k <= min(points.length, 10) The first k points are pairwise distinct. -10000 <= points[i][j] <= 10000
Reported by candidates. Source: FastPrep
Pattern and pitfall
The algorithm is plain simulation. Keep centroids as arrays. Each round, compare every point against every centroid using squared distance, so you never need a square root. Pick the smallest distance and break ties by the lower index, which a strict less-than comparison in index order gives you for free. Then recompute means per nonempty cluster. The pitfalls: an empty cluster must keep its old centroid, not reset to zero or divide by zero. Means can be fractional, so don't use integer division. Store sums as doubles or compare using scaled integers. Stop when no assignment changes, and check that on the assignment array, not the centroids. With 200 points, 10 dimensions and 10 clusters, brute force is fast. If the tie rule or the empty-cluster case slips your mind mid-assessment, StealthCoder is the hedge that catches it.
Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.
You can drill Vectorized K-Means Assignments cold, or you can hedge it. StealthCoder runs invisibly during screen share and surfaces a working solution in under 2 seconds. The proctor sees the IDE. They don't see what's behind it. Made by an engineer who treats the OA as theater. If yours is tonight, you don't have time to grind. You have time to hedge.
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Vectorized K-Means Assignments FAQ
What's the trick in the SambaNova K-Means question?+
There's no clever algorithm. The trick is following the spec exactly: first k points as centroids, squared distance, lowest index wins ties, and empty clusters keep their old centroid. Most failures come from missing one of those rules, not from the loop itself.
Why use squared distance instead of Euclidean distance?+
Squared distance preserves the ordering of distances, so the nearest centroid is the same either way. It skips the square root and avoids floating point error in comparisons. That matters for ties, because exact equality checks only stay reliable if you avoid sqrt.
How do I handle fractional centroids?+
Centroid means are usually non-integer, so store them as doubles. Compute the sum of each coordinate over the cluster, then divide by the cluster size. Don't use integer division. Squared distances to those centroids then compare correctly as doubles.
Could the loop run forever?+
Standard K-Means with deterministic tie-breaking converges, since each step doesn't increase total squared distance and assignments are discrete. Stop when a full pass changes no assignment. Compare the new assignment array against the previous one each iteration.
How do I prepare for this in 48 hours?+
Write the loop once from scratch, then test three cases: a tie like [[0],[4],[2]] with k=2, k equal to the number of points, and a cluster that goes empty. If those pass, the logic is solid. Spend the rest of the time on clean array handling.