703. Kth Largest Element in a Stream

Problem

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

  Constraints:

• 1 <= k <= 10^4

• 0 <= nums.length <= 10^4

• -10^4 <= nums[i] <= 10^4

• -10^4 <= val <= 10^4

• At most 10^4 calls will be made to add.

• It is guaranteed that there will be at least k elements in the array when you search for the kth element.

Solution (Java)

class KthLargest {
    private final int maxSize;
    private final PriorityQueue<Integer> heap;

    public KthLargest(int k, int[] nums) {
        this.maxSize = k;
        this.heap = new PriorityQueue<>();
        for (int num : nums) {
            this.add(num);
        }
    }

    public int add(int val) {
        if (heap.size() < maxSize) {
            heap.add(val);
        } else if (heap.peek() < val) {
            heap.add(val);
            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);
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).