Keys 23, 34 and 45 are inserted in that order into a hash table of size 11 using h(k) = k mod 11 and quadratic probing, h(k, i) = (h(k) + i²) mod 11. At which index is 45 stored?
Correct answer A. 5
Explanation
All three keys hash to 1. 23 takes index 1. 34 tries 1, then 1 + 1² = 2, which is free. 45 tries 1, then 2 (both occupied), then 1 + 2² = 5, which is free. Linear probing would have placed 45 at 3, so quadratic probing jumps further away, which reduces primary clustering.
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