703. 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 integer k and the stream of integers nums.
  • int add(int val) Appends the integer val to the stream and returns the element representing the kth largest element in the stream.

class KthLargest:
   
    def __init__(self, k: int, nums: List[int]):
        self.k = k
        self.heap = nums
        heapq.heapify(self.heap)
       
        while len(self.heap) > k:
            heapq.heappop(self.heap)

    def add(self, val: int) -> int:
        heapq.heappush(self.heap, val)
        if len(self.heap) > self.k:
            heapq.heappop(self.heap)

        return self.heap[0] 

Comments

Post a Comment

Popular posts from this blog

849. Maximize Distance to Closest Person

You are given an array representing a row of  seats  where  seats[i] = 1  represents a person sitting in the  i th  seat, and  seats[i] = 0  represents that the  i th  seat is empty  (0-indexed) . There is at least one empty seat, and at least one person sitting. Alex wants to sit in the seat such that the distance between him and the closest person to him is maximized.  Return  that maximum distance to the closest person . def maxDistToClosest ( self , seats : List [ int ]):                     taken = []         for i , occupied in enumerate ( seats ):         if occupied :             taken . append ( i )         S = len ( seats )     T = len ( taken )                 maxDist = max ( taken [ 0 ], S - taken [- 1 ] - 1 ) ...
Read more

347. Top K Frequent Elements

Given an integer array  nums  and an integer  k , return  the   k   most frequent elements . You may return the answer in  any order .   Example: Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2] def topKFrequent ( self , nums : List [ int ], k : int ):             freq = Counter ( nums )         return heapq . nlargest ( k , freq . keys (), key = freq . get )    
Read more

139. Word Break

Given a string  s  and a dictionary of strings  wordDict , return  true  if  s  can be segmented into a space-separated sequence of one or more dictionary words. Note  that the same word in the dictionary may be reused multiple times in the segmentation.   Example: Input: s = "leetcode", wordDict = ["leet","code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code".   def wordBreak ( s : str , wordDict : List [ str ]):             word_set = set ( wordDict )     dp = [ False ] * ( len ( s ) + 1 )     dp [ 0 ] = True     for i in range ( 1 , len ( s ) + 1 ):         for j in range ( i ):             if dp [ j ] and s [ j : i ] in word_set :                 dp [ i ] = True                 break   ...
Read more