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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
104
calls will be made to add
.k
elements in the array when you search for the kth
element. class KthLargest {
final PriorityQueue<Integer> q;
final int k;
public KthLargest(int k, int[] a) {
this.k = k;
q = new PriorityQueue<>(k);
for (int n : a)
add(n);
}
public int add(int n) {
if (q.size() < k)
q.offer(n);
else if (q.peek() < n) {
q.poll();
q.offer(n);
}
return q.peek();
}
}
class KthLargest(object):
def __init__(self, k, nums):
self.pool = nums
self.k = k
heapq.heapify(self.pool)
while len(self.pool) > k:
heapq.heappop(self.pool)
def add(self, val):
if len(self.pool) < self.k:
heapq.heappush(self.pool, val)
elif val > self.pool[0]:
heapq.heapreplace(self.pool, val)
return self.pool[0]
priority_queue<int, vector<int>, greater<int>> pq;
int size;
KthLargest(int k, vector<int> nums) {
size=k;
for(int i=0;i<nums.size();i++) {
pq.push(nums[i]);
if(pq.size()>k) pq.pop();
}
}
int add(int val) {
pq.push(val);
if(pq.size()>size) pq.pop();
return pq.top();
}
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