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