703. Kth Largest Element in a Stream
Description
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 integersnums
.int add(int val)
Appends the integerval
to the stream and returns the element representing thekth
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 toadd
. - It is guaranteed that there will be at least
k
elements in the array when you search for thekth
element.
Solution
kth-largest-element-in-a-stream.py
class KthLargest:
def __init__(self, k: int, nums: List[int]):
self.heap = []
self.k = k
for x in nums:
if len(self.heap) == k:
heapq.heappushpop(self.heap, x)
else:
heapq.heappush(self.heap, x)
def add(self, val: int) -> int:
if len(self.heap) == self.k:
heapq.heappushpop(self.heap, val)
else:
heapq.heappush(self.heap, val)
return self.heap[0]
# Your KthLargest object will be instantiated and called as such:
# obj = KthLargest(k, nums)
# param_1 = obj.add(val)