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I sorted the array first which was the right instinct, but then I fumbled around trying to brute force placements before realizing binary search on the answer was the cleaner path.
Recognize this as a minimax facility location problem on a line (1D) with L1 distance, which reduces to placing k centers to minimize the maximum gap between consecutive centers. Use binary search on the answer D, and for each D, greedily check if k centers can cover all points such that every point is within D of some center. The minimal feasible D is the answer.
Pro tip: Clarify that L1 distance in 1D is just absolute difference, so the problem simplifies to covering points with intervals of length 2D; this shows you can reduce complexity and avoid overcomplicating with general clustering algorithms.
Confirm that the array is 1D and distance is L1 (absolute difference). Restate the goal: minimize the maximum distance from any point to its nearest of k centers.
Observe that if a distance D is feasible, any larger D is also feasible. Binary search the minimal D over the range [0, max-min].
For a given D, sort the points and greedily place a center at the leftmost uncovered point + D, covering all points within D. Count centers needed; feasible if ≤ k.
Analyze time: O(n log n + n log(max-min)) due to sorting and binary search. Handle edge cases: k ≥ n (answer 0), k=1 (answer (max-min)/2), duplicate points.
Mention alternative approaches (e.g., DP for exact k centers) and why greedy+binary search is optimal and simpler for this minimax objective.
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