Solution
Correct Answer: (E)
\[ \text{constant} \cdot |S|(1 + \log |R|) \]
Key Observation
Recall the partition method used in Quicksort. The partition procedure places a chosen pivot element in its correct sorted position. If the partition returns an index \(k\), then the pivot element is exactly the \(k\)-th smallest element in the array.
Now consider the procedure \(\textsf{Select}(S, r)\). This procedure returns the element of rank \(r\) and also ensures that this element is placed in its correct position, with all elements smaller than it on the left and all larger elements on the right. Even if an implementation does not explicitly rearrange the array, this rearrangement can be done without affecting the asymptotic number of comparisons.
Hence, \(\textsf{Select}\) behaves in the same way as a partition step.
Objective
Efficiently implement \(\textsf{MultiSelect}(S, R)\), where \[ R = \{ r_1 < r_2 < \cdots < r_k \}. \]
Divide-and-Conquer Strategy
To understand the approach clearly, consider the following example: \[ R = \{2, 4, 6, 8, 10, 12\}. \] This means we want to find the 2nd, 4th, 6th, 8th, 10th, and 12th smallest elements of the set \(S\).
- First, choose the middle rank from \(R\). In this example, the middle rank is \(8\). Compute \(\textsf{Select}(S, 8)\), which finds the 8th smallest element using \(O(|S|)\) comparisons.
- After this operation, the 8th smallest element is placed in its correct position. All elements smaller than it lie to its left, and all elements larger than it lie to its right.
- The remaining ranks now split naturally around 8. The ranks \(\{2, 4, 6\}\) correspond to elements that must lie in the left subarray, while the ranks \(\{10, 12\}\) correspond to elements that must lie in the right subarray.
- Next, apply the same idea recursively on both sides. To find the 4th smallest element, apply \(\textsf{Select}\) only on the left subarray. To find the 10th smallest element, apply \(\textsf{Select}\) only on the right subarray. Even though two selection operations are performed, the total number of comparisons at this level is still \(O(|S|)\).
- After locating the 4th, 8th, and 10th smallest elements, the array is divided into four parts:
- Elements smaller than the 4th smallest element
- Elements between the 4th and 8th smallest elements
- Elements between the 8th and 10th smallest elements
- Elements larger than the 10th smallest element
- The same process is applied recursively to each part that contains at least one required rank. At each level of recursion, the number of selected ranks doubles, while the total work done at that level remains proportional to \(|S|\).
Cost Analysis
At each level of recursion, the total number of comparisons is \(O(|S|)\). The number of levels required to process all \(|R|\) ranks is: \[ \log |R| + 1. \]
Therefore, the total number of comparisons is: \[ O(|S|) \times (\log |R| + 1) = \text{constant} \cdot |S|(1 + \log |R|). \]
Final Answer
\[ \boxed{\text{constant} \cdot |S|(1 + \log |R|)} \]
Hence, the correct option is (E).
A similar question can be found on page 6, question number 5, available at the following link: practicefinalsol.pdf .
