1. **Problem Statement:** We have $n$ items to sort using BucketSort with $k = \frac{n}{\log \log n}$ buckets. The input distribution is non-uniform: half the items are uniformly distributed in $[0, \frac{1}{3}]$ and the other half in $(\frac{1}{3}, 1]$. We use heap-sort to sort items within each bucket. We want to find the expected running time. 2. **Recall BucketSort running time:** - BucketSort divides the interval into $k$ buckets. - Items are distributed into buckets. - Each bucket is sorted individually. - Total time is $O(n + \sum_{i=1}^k T_i)$ where $T_i$ is the time to sort bucket $i$. 3. **Distribution of items per bucket:** - Since half the items are in $[0, \frac{1}{3}]$ and half in $(\frac{1}{3}, 1]$, the buckets covering $[0, \frac{1}{3}]$ get about $\frac{n}{2}$ items spread over roughly $\frac{k}{3}$ buckets. - Buckets in $[0, \frac{1}{3}]$ have expected bucket size $\approx \frac{n/2}{k/3} = \frac{3n}{2k}$. - Buckets in $(\frac{1}{3}, 1]$ get the other half $\frac{n}{2}$ items over $\frac{2k}{3}$ buckets, so expected bucket size $\approx \frac{n/2}{2k/3} = \frac{3n}{4k}$. 4. **Substitute $k = \frac{n}{\log \log n}$:** - Bucket size in $[0, \frac{1}{3}]$ is $\frac{3n}{2 \cdot n/(\log \log n)} = \frac{3}{2} \log \log n$. - Bucket size in $(\frac{1}{3}, 1]$ is $\frac{3n}{4 \cdot n/(\log \log n)} = \frac{3}{4} \log \log n$. 5. **Sorting time per bucket using heap-sort:** - Heap-sort time for a bucket of size $m$ is $O(m \log m)$. - For buckets in $[0, \frac{1}{3}]$, time per bucket is $O(\log \log n \cdot \log(\log \log n))$. - For buckets in $(\frac{1}{3}, 1]$, time per bucket is similar order. 6. **Number of buckets:** - Total buckets $k = \frac{n}{\log \log n}$. 7. **Total sorting time inside buckets:** - Sum over all buckets: $k \times O(\log \log n \cdot \log(\log \log n)) = \frac{n}{\log \log n} \cdot \log \log n \cdot \log(\log \log n) = n \log(\log \log n)$. 8. **Total expected running time:** - BucketSort overhead is $O(n)$ for distribution. - Sorting inside buckets is $O(n \log(\log \log n))$. - Dominant term is $O(n \log(\log \log n))$. 9. **Compare with given options:** - None of the options exactly match $\theta(n \log(\log \log n))$. - The closest is option (k) None of the other choices are correct. **Final answer:** k. $\theta(n \log(\log \log n))$ is not listed, so answer is (k).