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.

Your KthLargest class will have a constructor which accepts an integer k and an integer array nums, which contains initial elements from the stream. For each call to the method KthLargest.add, return the element representing the kth largest element in the stream.

Example:

int k = 3;
int[] arr = [4,5,8,2];
KthLargest kthLargest = new KthLargest(3, arr);
kthLargest.add(3);   // returns 4
kthLargest.add(5);   // returns 5
kthLargest.add(10);  // returns 5
kthLargest.add(9);   // returns 8
kthLargest.add(4);   // returns 8

Note: 
You may assume that nums‘ length ≥ k-1 and k ≥ 1.

Solution: BST / Min Heap

Time complexity: O(nlogk)

Space complexity: O(k)

C++ / BST

// Author: Huahua
// Running time: 32 ms
class KthLargest {
public:
  KthLargest(int k, vector<int> nums): k_(k) {
    for (int num : nums)
      add(num);
  }

  int add(int val) {    
    s_.insert(val);
    if (s_.size() > k_)
      s_.erase(s_.begin());
    return *s_.begin();
  }
private:
  const int k_;  
  multiset<int> s_;
};

C++ / Min Heap

// Author: Huahua
// Running time: 28 ms
class KthLargest {
public:
  KthLargest(int k, vector<int> nums): k_(k) {    
    for (int num : nums)
      add(num);
  }

  int add(int val) {    
    s_.push(val);
    if (s_.size() > k_)
      s_.pop();
    return s_.top();
  }
private:
  const int k_;  
  priority_queue<int, vector<int>, greater<int>> s_; // min heap
};

The post 花花酱 LeetCode 703. Kth Largest Element in a Stream appeared first on Huahua's Tech Road.

Problem:

Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value.

Examples:

[2,3,4] , the median is 3

[2,3], the median is (2 + 3) / 2 = 2.5

Design a data structure that supports the following two operations:

• void addNum(int num) – Add a integer number from the data stream to the data structure.
• double findMedian() – Return the median of all elements so far.

For example:

addNum(1)
addNum(2)
findMedian() = 1.5
addNum(3) 
findMedian() = 2

Idea:

1. Min/Max heap
2. Balanced binary search tree

Time Complexity:

add(num): O(logn)

findMedian(): O(logn)

Solution1:

// Author: Huahua
// Running Time: 152 ms
class MedianFinder {
public:
    /** initialize your data structure here. */
    MedianFinder() {}
    
    // l_.size() >= r_.size()
    void addNum(int num) {
        if (l_.empty() || num <= l_.top()) {
            l_.push(num);
        } else {
            r_.push(num);
        }
        
        // Step 2: Balence left/right
        if (l_.size() < r_.size()) {
            l_.push(r_.top());
            r_.pop();
        } else if (l_.size() - r_.size() == 2) {
            r_.push(l_.top());
            l_.pop();
        }
    }
    
    double findMedian() {
        if (l_.size() > r_.size()) {
            return static_cast<double>(l_.top());
        } else {            
            return (static_cast<double>(l_.top()) + r_.top()) / 2;
        }
    }
private:
    priority_queue<int, vector<int>, less<int>> l_;    // max-heap
    priority_queue<int, vector<int>, greater<int>> r_; // min-heap
};

Solution 2:

// Author: Huahua
// Running time: 172 ms
class MedianFinder {
public:
    /** initialize your data structure here. */
    MedianFinder(): l_(m_.cend()), r_(m_.cend()) {}
    
    // O(logn)
    void addNum(int num) {
        if (m_.empty()) {
            l_ = r_ = m_.insert(num);
            return;
        }
        
        m_.insert(num);
        const size_t n = m_.size();    
        
        if (n & 1) {
            // odd number
            if (num >= *r_) {         
                l_ = r_;
            } else {
                // num < *r_, l_ could be invalidated
                l_ = --r_;
            }
        } else {
            if (num >= *r_)
                ++r_;
            else
                --l_;
        }
    }
    // O(1)
    double findMedian() {
        return (static_cast<double>(*l_) + *r_) / 2;
    }
private:
    multiset<int> m_;
    multiset<int>::const_iterator l_;  // current left median
    multiset<int>::const_iterator r_;  // current right median
};

Related Problems

• 花花酱 LeetCode 480. Sliding Window Median
• [解题报告] LeetCode 4. Median of Two Sorted Arrays
• [ZOJ] 3612: Median

The post 花花酱 LeetCode 295. Find Median from Data Stream O(logn) + O(1) appeared first on Huahua's Tech Road.