kth-largest-element-in-a-stream.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
//
//
//  
// Constraints:
//
//
// 	1 <= k <= 104
// 	0 <= nums.length <= 104
// 	-104 <= nums[i] <= 104
// 	-104 <= val <= 104
// 	At most 104 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.
//
//


class KthLargest {
    private PriorityQueue<Integer> q = new PriorityQueue<>();
    private int k;
    public KthLargest(int k, int[] nums) {
        this.k = k;
        for (int num : nums) {
            add(num);
        }
    }
    
    public int add(int val) {
        q.add(val);
        if (q.size() > k) {
            q.poll();
        }
        return q.peek();
    }
}

/**
 * Your KthLargest object will be instantiated and called as such:
 * KthLargest obj = new KthLargest(k, nums);
 * int param_1 = obj.add(val);
 */