Description
https://leetcode.com/problems/kth-largest-element-in-a-stream/
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)
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.
Explanation
A simple solution is to sort the list whenever adding a new value and then find the kth largest value.
Python Solution
class KthLargest:
def __init__(self, k: int, nums: List[int]):
self.nums = nums
self.k = k
def add(self, val: int) -> int:
self.nums.append(val)
self.nums = sorted(self.nums)
return self.nums[len(self.nums) - self.k]
# Your KthLargest object will be instantiated and called as such:
# obj = KthLargest(k, nums)
# param_1 = obj.add(val)
- Time Complexity: O(NlogN).
- Space Complexity: O(N).