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) 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 {
int[] pq; // k + 1
int n = 0;
int k;
public KthLargest(int k, int[] nums) {
this.k = k;
this.pq = new int[k + 1];
for (int i = 0; i < nums.length; i++) {
add(nums[i]);
}
}
public int add(int val) {
if (n < k) {
pq[++n] = val;
swim(n);
} else {
if (val > pq[1]) {
pq[1] = val;
sink(1);
}
}
return pq[1];
}
void swim(int k) {
while (k > 1 && !less(k/2, k)) {
exch(k/2, k);
k = k / 2;
}
}
void sink(int k) {
while (2 * k < pq.length) {
int j = 2 * k;
if (j < pq.length - 1 && !less(j, j+ 1)) j++;
if (less(k, j)) break;
exch(k, j);
k = j;
}
}
int delMin() {
int max = pq[1];
exch(1, n--);
sink(1);
pq[n+1] = -1;
return max;
}
boolean isEmpty() {
return n == 0;
}
boolean less(int i, int j) {
return pq[i] < pq[j];
}
void exch(int i, int j) {
int t = pq[i];
pq[i] = pq[j];
pq[j] = t;
}
}
/**
* Your KthLargest object will be instantiated and called as such:
* KthLargest obj = new KthLargest(k, nums);
* int param_1 = obj.add(val);
*/
Solution 2021-11-20
class KthLargest {
PriorityQueue<Integer> q = new PriorityQueue<>();
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);
*/
Notes:
- Ask about unique value
- Ex1: 6,5,4,2 and k = 3 will return 4
- Ex2: 6,5,5,2 and k = 3 will return 4 or 5?