Tag
: Heap
Design
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 integerk
and the stream of integers nums.int add(int val)
Appends the integerval
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:
def __init__(self, k: int, nums: List[int]):
self.k = k
self.h = nums
heapq.heapify(self.h)
while len(self.h) > k:
heapq.heappop(self.h)
def add(self, val: int) -> int:
heapq.heappush(self.h, val)
if len(self.h) > self.k:
heapq.heappop(self.h)
return self.h[0]
# Your KthLargest object will be instantiated and called as such:
# obj = KthLargest(k, nums)
# param_1 = obj.add(val)