流中的第K个最大元素

给定无限个整数流，请在任何时间点找到第k个最大元素。
例子：

Input:
stream[] = {10, 20, 11, 70, 50, 40, 100, 5, ...}
k = 3

Output:    {_,   _, 10, 11, 20, 40, 50,  50, ...}

允许的额外空间为O(k)。

在以下文章中，我们讨论了在数组中找到第k个最大元素的不同方法。
未排序数组中第K个最小/最大元素|套装1
未排序数组中第K个最小/最大元素|设置2(预期线性时间)
未排序数组中第K个最小/最大元素|组合3(最坏情况的线性时间)
在这里，我们有一个流而不是整个数组，并且只允许存储k个元素。

一个简单的解决方案是保持大小为k的数组。这个想法是保持数组排序，以便在O(1)时间内找到第k个最大元素(如果数组按升序排序，我们只需要返回数组的第一个元素)
如何处理流的新元素?
对于流中的每个新元素，检查新元素是否小于当前第k个最大元素。如果是，则忽略它。如果否，则从数组中删除最小的元素，并按排序顺序插入新元素。处理新元素的时间复杂度为O(k)。

更好的解决方案是使用大小为k的自平衡二进制搜索树。可以在O(Logk)时间中找到第k个最大元素。
如何处理流的新元素?
对于流中的每个新元素，检查新元素是否小于当前第k个最大元素。如果是，则忽略它。如果否，则从树中删除最小的元素并插入新元素。处理新元素的时间复杂度为O(Logk)。

一种有效的解决方案是使用大小为k的最小堆存储k个流的最大元素。第k个最大元素始终位于根，可以在O(1)时间找到。
如何处理流的新元素?
将新元素与堆根进行比较。如果新元素较小，则将其忽略。否则，将root替换为新元素，并调用heapify作为修改后的堆的根。找到第k个最大元素的时间复杂度是O(Logk)。

CPP

// A C++ program to find k'th
// smallest element in a stream
#include 
#include 
using namespace std;
 
// Prototype of a utility function
// to swap two integers
void swap(int* x, int* y);
 
// A class for Min Heap
class MinHeap {
    int* harr; // pointer to array of elements in heap
    int capacity; // maximum possible size of min heap
    int heap_size; // Current number of elements in min heap
public:
    MinHeap(int a[], int size); // Constructor
    void buildHeap();
    void MinHeapify(
        int i); // To minheapify subtree rooted with index i
    int parent(int i) { return (i - 1) / 2; }
    int left(int i) { return (2 * i + 1); }
    int right(int i) { return (2 * i + 2); }
    int extractMin(); // extracts root (minimum) element
    int getMin() { return harr[0]; }
 
    // to replace root with new node x and heapify() new
    // root
    void replaceMin(int x)
    {
        harr[0] = x;
        MinHeapify(0);
    }
};
 
MinHeap::MinHeap(int a[], int size)
{
    heap_size = size;
    harr = a; // store address of array
}
 
void MinHeap::buildHeap()
{
    int i = (heap_size - 1) / 2;
    while (i >= 0) {
        MinHeapify(i);
        i--;
    }
}
 
// Method to remove minimum element
// (or root) from min heap
int MinHeap::extractMin()
{
    if (heap_size == 0)
        return INT_MAX;
 
    // Store the minimum vakue.
    int root = harr[0];
 
    // If there are more than 1 items,
    // move the last item to
    // root and call heapify.
    if (heap_size > 1) {
        harr[0] = harr[heap_size - 1];
        MinHeapify(0);
    }
    heap_size--;
 
    return root;
}
 
// A recursive method to heapify a subtree with root at
// given index This method assumes that the subtrees are
// already heapified
void MinHeap::MinHeapify(int i)
{
    int l = left(i);
    int r = right(i);
    int smallest = i;
    if (l < heap_size && harr[l] < harr[i])
        smallest = l;
    if (r < heap_size && harr[r] < harr[smallest])
        smallest = r;
    if (smallest != i) {
        swap(&harr[i], &harr[smallest]);
        MinHeapify(smallest);
    }
}
 
// A utility function to swap two elements
void swap(int* x, int* y)
{
    int temp = *x;
    *x = *y;
    *y = temp;
}
 
// Function to return k'th largest element from input stream
void kthLargest(int k)
{
    // count is total no. of elements in stream seen so far
    int count = 0, x; // x is for new element
 
    // Create a min heap of size k
    int* arr = new int[k];
    MinHeap mh(arr, k);
 
    while (1) {
        // Take next element from stream
        cout << "Enter next element of stream ";
        cin >> x;
 
        // Nothing much to do for first k-1 elements
        if (count < k - 1) {
            arr[count] = x;
            count++;
        }
 
        else {
            // If this is k'th element, then store it
            // and build the heap created above
            if (count == k - 1) {
                arr[count] = x;
                mh.buildHeap();
            }
 
            else {
                // If next element is greater than
                // k'th largest, then replace the root
                if (x > mh.getMin())
                    mh.replaceMin(x); // replaceMin calls
                                      // heapify()
            }
 
            // Root of heap is k'th largest element
            cout << "K'th largest element is "
                 << mh.getMin() << endl;
            count++;
        }
    }
}
 
// Driver program to test above methods
int main()
{
    int k = 3;
    cout << "K is " << k << endl;
    kthLargest(k);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG {
   
  /*
  using min heap DS
 
  how data are stored in min Heap DS
         1
       2   3
  if k==3 , then top elemnt of heap
  itself the kth largest largest element
 
  */
  static PriorityQueue min;
  static int k;
 
  static List getAllKthNumber(int arr[])
  {
     
    // list to store kth largest number
    List list = new ArrayList<>();
 
    // one by one adding values to the min heap
    for (int val : arr) {
 
      // if the heap size is less than k , we add to
      // the heap
      if (min.size() < k)
        min.add(val);
 
      /*
      otherwise ,
      first we  compare the current value with the
      min heap TOP value
 
      if TOP val > current element , no need to
      remove TOP , bocause it will be the largest kth
      element anyhow
 
      else  we need to update the kth largest element
      by removing the top lowest element
      */
 
      else {
        if (val > min.peek()) {
          min.poll();
          min.add(val);
        }
      }
 
      // if heap size >=k we add
      // kth largest element
      // otherwise -1
 
      if (min.size() >= k)
        list.add(min.peek());
      else
        list.add(-1);
    }
    return list;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    min = new PriorityQueue<>();
 
    k = 4;
 
    int arr[] = { 1, 2, 3, 4, 5, 6 };
 
    List res = getAllKthNumber(arr);
 
    for (int x : res)
      System.out.print(x + " ");
  }
 
    // This code is Contributed by Pradeep Mondal P
}

输出：

K is 3
Enter next element of stream 23
Enter next element of stream 10
Enter next element of stream 15
K'th largest element is 10
Enter next element of stream 70
K'th largest element is 15
Enter next element of stream 5
K'th largest element is 15
Enter next element of stream 80
K'th largest element is 23
Enter next element of stream 100
K'th largest element is 70
Enter next element of stream
CTRL + C pressed

使用优先级队列的实施：

C++

// C++ program for the above approach
#include 
using namespace std;
 
vector kthLargest(int k, int arr[], int n)
{
    vector ans(n);
 
    // Creating a min-heap using priority queue
    priority_queue, greater > pq;
 
    // Iterating through each element
    for (int i = 0; i < n; i++) {
        // If size of priority
        // queue is less than k
        if (pq.size() < k)
            pq.push(arr[i]);
        else {
            if (arr[i] > pq.top()) {
                pq.pop();
                pq.push(arr[i]);
            }
        }
 
        // If size is less than k
        if (pq.size() < k)
            ans[i] = -1;
        else
            ans[i] = pq.top();
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    int n = 6;
    int arr[n] = { 1, 2, 3, 4, 5, 6 };
    int k = 4;
   
    // Function call
    vector v = kthLargest(k, arr, n);
    for (auto it : v)
        cout << it << " ";
    return 0;
}

输出

-1 -1 -1 1 2 3