# 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)