Sample test driver code: int[] myArray = (7, 4, 6, 10, 9, 1, 20); int k = 4; System.out.println("The " + findKthSmallest (myArray, k)); + k + "th smallest element is: Sample Output: The 4th smallest element is:7
Sample test driver code: int[] myArray = (7, 4, 6, 10, 9, 1, 20); int k = 4; System.out.println("The " + findKthSmallest (myArray, k)); + k + "th smallest element is: Sample Output: The 4th smallest element is:7
Related questions
Question
![Question 2: A common task in day-to-day operations is to find the k-th smallest element in a given
data structure, be it an array or a list. A typical way of completing this task is to sort the array and
then get the k-th element. The former takes at least O(nlogn) time, where n is the size of the array,
and the latter takes constant time. Therefore, the total complexity is O(nlogn).
A better solution is to take advantage of heaps, which takes O(logn) time to insert an element. To
simplify your implementation, you will use the Java built-in priority queue (with heaps as the
underlying data structure) to find the k-th smallest element in a given array. The general idea is to
build a heap for the first k elements in the array and then for the rest of the (n - k) elements, compare
each of them with the root of the heap (i.e., the largest number in the heap) - if the element is
smaller than the root, remove the root and insert the element into the heap. Repeat this process
until the array is exhausted. In the end, your heap will hold the smallest k elements with the root
being the largest - it is also the k-th smallest of the array. Since the height of the heap is always
kept at k, inserting an element takes no more than O(logk) time. In the worst-case scenario, you
would end up comparing and inserting (n- k) elements into the heap while maintaining its height.
Accordingly, the total operation takes no more than O(nlogk) time- faster than O(nlogn).
Step 1 - Create the method findKthSmallest which takes as parameters an int array
numArray and an int k.
The Java built-in priority queue implementation prioritizes the smallest element in the
queue. However, we could invoke the method reverseOrder from the interface
Comparator to reverse the priority.
You can use a PriorityQueue to store elements that are smaller than or equal to the
k-th smallest element by traversing numArray starting after the k-th index to find the
smallest elements.
The k-th smallest element should be first in the queue if we reverse the order of the
PriorityQueue.
Step 2- Create a test driver to verify your implementation and use the jGRASP plugin to visualize
how the priority queue evolves.
Sample test driver code:
int[] myArray = (7, 4, 6, 10, 9, 1, 20);
int x = 4;
" + k
System.out.println ("The
findKthSmallest (myArray, k));
"th
smallest element is:
Sample Output:
The 4th smallest element is: 7](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd36b864c-a7e3-4195-bc2c-e4ce0ce4787e%2F5ecea658-b3fd-4b41-be04-3d4e242f0f13%2Fkzf1v4o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 2: A common task in day-to-day operations is to find the k-th smallest element in a given
data structure, be it an array or a list. A typical way of completing this task is to sort the array and
then get the k-th element. The former takes at least O(nlogn) time, where n is the size of the array,
and the latter takes constant time. Therefore, the total complexity is O(nlogn).
A better solution is to take advantage of heaps, which takes O(logn) time to insert an element. To
simplify your implementation, you will use the Java built-in priority queue (with heaps as the
underlying data structure) to find the k-th smallest element in a given array. The general idea is to
build a heap for the first k elements in the array and then for the rest of the (n - k) elements, compare
each of them with the root of the heap (i.e., the largest number in the heap) - if the element is
smaller than the root, remove the root and insert the element into the heap. Repeat this process
until the array is exhausted. In the end, your heap will hold the smallest k elements with the root
being the largest - it is also the k-th smallest of the array. Since the height of the heap is always
kept at k, inserting an element takes no more than O(logk) time. In the worst-case scenario, you
would end up comparing and inserting (n- k) elements into the heap while maintaining its height.
Accordingly, the total operation takes no more than O(nlogk) time- faster than O(nlogn).
Step 1 - Create the method findKthSmallest which takes as parameters an int array
numArray and an int k.
The Java built-in priority queue implementation prioritizes the smallest element in the
queue. However, we could invoke the method reverseOrder from the interface
Comparator to reverse the priority.
You can use a PriorityQueue to store elements that are smaller than or equal to the
k-th smallest element by traversing numArray starting after the k-th index to find the
smallest elements.
The k-th smallest element should be first in the queue if we reverse the order of the
PriorityQueue.
Step 2- Create a test driver to verify your implementation and use the jGRASP plugin to visualize
how the priority queue evolves.
Sample test driver code:
int[] myArray = (7, 4, 6, 10, 9, 1, 20);
int x = 4;
" + k
System.out.println ("The
findKthSmallest (myArray, k));
"th
smallest element is:
Sample Output:
The 4th smallest element is: 7
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 4 steps with 4 images
