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703. Kth Largest Element in a Stream

https://leetcode.com/problems/kth-largest-element-in-a-stream/

Level: Easy

Related Topic: Heap

Related Problem:

Heap Notes

Solution​

Trivial Solution with heap (will timeout)​

Since python's heapq library implements a min heap, I store all the numbers after inverting them (times -1).

When adding a new number, I push the value's opposite value first, and pop k values.

The last popped value is the k-th largest which will be returned. Then push everything back to the queue.

This will timeout!

Support len(nums) == N, number of times add is called = M, k is the k-th max value to return.

class KthLargest:

    def __init__(self, k: int, nums: List[int]):
        self.heap = [-x for x in nums]
        heapify(self.heap)				# make max heap
        self.k = k

    def add(self, val: int) -> int:
        heappush(self.heap, -val)
        backup = [heappop(self.heap) for i in range(self.k)]
        for x in backup:
            heappush(self.heap, x)
        return -backup[-1]

Faster Heap Solution with a Trick​

Time Complexity:​

O(Nlog(N) + Mlog(k)) time in total.

  • Heapify takes O(N) time in python.
  • Each pop or push to heap takes O(log(N)) time.
    • If k==N, it takes at most O(N log(N)) to pop
  • __init__ takes O(N + N log(N)) time
  • add function push once and pop once, which takes O(2 * log(N)) time
  • M add costs O(M*log(k)) time

Space Complexity:​

Costs O(N + M) to store a heap (regular array) of size N+M (initial size N, and M added elements).

Read Heap Notes first.

Runtime: 111 ms, faster than 76.78% of Python3 online submissions for Kth Largest Element in a Stream.

Memory Usage: 18.5 MB, less than 10.77% of Python3 online submissions for Kth Largest Element in a Stream.

class KthLargest:

    def __init__(self, k: int, nums: List[int]):
        self.heap = nums
        heapify(self.heap)      # make min heap
        self.k = k
        while len(self.heap) > k:
            # when there are k elements left in a min heap, the next popped value should be the k-th max value
			# the popped items are all smaller than the k-th max, and could never become the k-th max.
			# Because elements are added (no removing), the k-th max could only get larger and larger.
            heappop(self.heap)

    def add(self, val: int) -> int:
		# After pushing an item and popping one, the next element is still the k-th max.
        heappush(self.heap, val)
        if len(self.heap) > self.k:
			# this conditional statement is for avoiding empty queue
            heappop(self.heap)
        # a heap array's index 0 is the next item to be popped, saves time from popping and pushing again
        return self.heap[0]


# Your KthLargest object will be instantiated and called as such:
# obj = KthLargest(k, nums)
# param_1 = obj.add(val)

Super Short Solution with SortedList​

Runtime: 147 ms, faster than 47.73% of Python3 online submissions for Kth Largest Element in a Stream.

Memory Usage: 18.7 MB, less than 6.16% of Python3 online submissions for Kth Largest Element in a Stream.

Not sure how SortedList is implemented, but whatever is added to the sorted list, it's always in order.

I guess data is stored in a regular list structure to support fastest access (O(1)) with index.

In this case, adding a new element takes at least O(log(N)) time. O(N) to perform insertion sort style insertion, or O(log(N)) to perform binary earch, and O(N) to shift elements.

Time Complexity: O(N*log(N) + M * N)

Space Complexity: O(N + M) to store a regular array.

In other languages without SortedList library, we have to impelement a binary search and insertion manually.

See JavaScript Example.

from sortedcontainers import SortedList

class KthLargest:

    def __init__(self, k, nums):
        self.k, self.sl = k, SortedList(nums)

    def add(self, val):
        self.sl.add(val)  # Note that sl is a SortedList
        return self.sl[-self.k]  # Note that sl is in ascending order