46 lines
1.2 KiB
Python
46 lines
1.2 KiB
Python
# Naive approach, will TLE
|
|
# Time for each add: O(k)
|
|
class KthLargest:
|
|
|
|
def __init__(self, k: int, nums: List[int]):
|
|
self.k = k
|
|
|
|
len_nums = len(nums)
|
|
|
|
# Only keep max k elements, rest are useless
|
|
if len_nums >= k:
|
|
self.nums = sorted(nums)[len_nums - k:]
|
|
# If we're provided less than k elements, we
|
|
# will it up with least possible number
|
|
else:
|
|
self.nums = [
|
|
*([float("-inf")] * (k - len_nums)),
|
|
*sorted(nums)
|
|
]
|
|
|
|
def add(self, val: int) -> int:
|
|
# Find the right place for val in the list
|
|
# of length k
|
|
|
|
i = self.k - 1
|
|
while i >= 0 and val < self.nums[i]:
|
|
i -= 1
|
|
|
|
# If val is big enough to fit in the list, shift
|
|
# existing elements to left
|
|
j = 0
|
|
while j < i:
|
|
self.nums[j] = self.nums[j + 1]
|
|
j += 1
|
|
|
|
if i >= 0:
|
|
self.nums[i] = val
|
|
|
|
# kth largest element is always at index 0
|
|
return self.nums[0]
|
|
|
|
|
|
# Your KthLargest object will be instantiated and called as such:
|
|
# obj = KthLargest(k, nums)
|
|
# param_1 = obj.add(val)
|