study/java

Design a class to find the kth largest element in a stream. Note that it is the kth largest element in the sorted order, not the kth distinct element.

Implement KthLargest class:

• KthLargest(int k, int[] nums) Initializes the object with the integer k and the stream of integers nums.
• int add(int val) Appends the integer val to the stream and returns the element representing the kth largest element in the stream.

Example 1:

Input
["KthLargest", "add", "add", "add", "add", "add"]
[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]
Output
[null, 4, 5, 5, 8, 8]

Explanation
KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
kthLargest.add(3); // return 4
kthLargest.add(5); // return 5
kthLargest.add(10); // return 5
kthLargest.add(9); // return 8
kthLargest.add(4); // return 8

Solution

class KthLargest {

  private PriorityQueue<Integer> heap = new PriorityQueue<>();

  private int k;
  

  public KthLargest(int k, int[] nums) {

    this.k = k;

    for (var n : nums) add(n);

  }

  public int add(int val) {

    heap.offer(val);

    if (heap.size() > k) heap.poll();

    return heap.peek();

  }

}

  

/**

 * Your KthLargest object will be instantiated and called as such:

 * KthLargest obj = new KthLargest(k, nums);

 * int param_1 = obj.add(val);

 */

PriorityQueue

• add (E e)
• clear ()
• comparator ()
• contains (Object o)
• iterator ()
• offer (E e)
• peek ()
• poll ()
• remove (Oject o)
• size ()
• spliterator ()
• toArray ()
• toArray (T[] a)