The K-th Element

The K-th Element - Approach 1
Use the Heap data structure to obtain the K-th largest or smallest element.

---

Solution of the K-th largest element:

1. Construct a Max Heap
2. Add all elements into the Max Heap
3. Traversing and deleting the top element (using pop() or poll() for instance)
4. Repeat Step 3 K times until we find the K-th largest element.

Solution of the K-th smallest element:

1. Construct a Min Heap.
2. Add all elements into the Min Heap
3. Traversing and deleting the top element (using pop() or poll() for instance)
4. Repeat Step 3 K times until we find the k-th smallest element.

---

Complexity Analysis:
Let N be the total number of elements.

Time complexity: O(K log N + N)

• Steps one and two require us to construct a Max Heap which requires O(N) time using the previously discussed heapify method. Each element removed from the heap requires O(log N) time; this process is repeated K times. Thus the total time complexity is O(K log N + N)

Space complexity: O(N)

• After step 2, the heap will store all N elements.

---

The K-th Element - Approach 2

Solution of the K-th largest element:

1. Construct a Min Heap with size K
2. Add elements to the Min Heap one by one
3. When there are K elements in the “Min Heap”, compare the current element with the top element of the Heap:
   • If the current element is not larger than the top element of the Heap, drop it and proceed to the next element
   • If the current element is larger than the Heap’s top element, delete the Heap’s top element, and add the current element to the “Min Heap”
4. Repeat Steps 2 and 3 until all elements have been iterated.
5. Now the top element in the Min Heap is the K-th largest element

Solution of the K-th smallest element:

1. Construct a Max Heap with size K.
2. Add elements to the Max Heap one by one
3. When there are K elements in the Max Heap, compare the current element with the top element of the Heap:
   • If the current element is not smaller than the top element of the Heap, drop it and proceed to the next element;
   • If the current element is smaller than the top element of the Heap, delete the top element of the Heap, and add the current element to the Max Heap.
4. Repeat Steps 2 and 3 until all elements have been iterated. Now the top element in the Max Heap is the K smallest element.

---

Complexity Analysis:
Time complexity: O(N log k)

• Steps one and two will require O(K log K) time if the elements are added one by one to the heap, however using the heapify method, these two steps could be accomplished in O(K) time. Steps 3 and 4 will require O(log K) time each time an element must be replaced in the heap. In the worst-case scenario, this will be done N - K times. Thus the total time complexity is O((N - K) log K + K log K) which simplifies to O(N log K).

Space complexity: O(K)

• The heap will contain at most K elements at any given time.