K-ésimo elemento más grande después de cada inserción

Dada una secuencia infinita de números enteros, encuentre el k-ésimo elemento más grande en cualquier punto del tiempo. Se puede suponer que 1 <= k <= n. 
 

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

El espacio extra permitido es O(k).
 

La idea es usar el montón mínimo. 
1) Almacene los primeros k elementos en el montón mínimo. 
2) Para cada elemento de (k+1)-ésimo a n-ésimo, haga lo siguiente. 
……a) Imprime la raíz del montón. 
……b) Si el elemento actual es más que la raíz del montón, extraiga la raíz e inserte
 

// CPP program to find k-th largest element in a
// stream after every insertion.
#include <bits/stdc++.h>
using namespace std;
 
int kthLargest(int stream[], int n, int k)
{
   // Create a min heap and store first k-1 elements
   // of stream into
   priority_queue<int, vector<int>, greater<int> > pq;
 
   // Push first k elements and print "_" (k-1) times
   for (int i=0; i<k-1; i++)
   {
      pq.push(stream[i]);
      cout << "_ ";
   }
   pq.push(stream[k-1]);
 
   for (int i=k; i<n; i++)
   {
       // We must insert last element before we
       // decide last k-th largest output.
          cout << pq.top() << " ";
 
       if (stream[i] > pq.top())
       {
           pq.pop();
           pq.push(stream[i]);
       } 
   } 
 
   // Print last k-th largest element (after
   // (inserting last element)
   cout << pq.top();
}
 
// Driver code
int main()
{
   int arr[] = {10, 20, 11, 70, 50, 40, 100, 55};
   int k = 3;
   int n = sizeof(arr)/sizeof(arr[0]);
   kthLargest(arr, n, k);
   return 0;
}

// Java Program for the above approach
import java.util.*;
class GFG
{
  static void kthLargest(int stream[], int n, int k)
  {
 
    // Create a min heap and store first k-1 elements
    // of stream into
    Vector<Integer> pq = new Vector<Integer>(n);
 
    // Push first k elements and print "_" (k-1) times
    for (int i = 0; i < k - 1; i++)
    {
      pq.add(stream[i]);
      System.out.print("_ ");
    }
    pq.add(stream[k - 1]);
 
    for (int i = k; i < n; i++)
    {
 
      // We must insert last element before we
      // decide last k-th largest output.
      Collections.sort(pq);
      System.out.print(pq.get(0) + " ");     
      if (stream[i] > pq.get(0))
      {
        pq.remove(0);
        pq.add(stream[i]);
      }  
    }  
 
    // Print last k-th largest element (after
    // (inserting last element)
    Collections.sort(pq);
    System.out.print(pq.get(0));
  }
 
  // Driver code
  public static void main(String[] args) {
    int arr[] = {10, 20, 11, 70, 50, 40, 100, 55};
    int k = 3;
    int n = arr.length;
    kthLargest(arr, n, k);
  }
}
 
// This code is contributed by divyeshrabadiya07.

## Python Program for the above approach
import heapq
def kthLargest(stream, n, k):
    # Create a min heap and store first k-1 elements
    # of stream into
    pq = []
    # Push first k elements and print "_" (k-1) times
    for i in range(k - 1):
        pq.append(stream[i])
        print("_ ", end = "")
    pq.append(stream[k - 1])
    # creating min heap and maintaining the heap
    # property
    heapq.heapify(pq)
     
    for i in range(k,n):
        # We must insert last element before we
        # decide last k-th largest output.
        print(pq[0],end=" ")
         
        if (stream[i]>pq[0]):
            #deleting the last element from the min heap
            pq[0]=pq[-1]
            pq.pop()
            pq.append(stream[i])
            heapq.heapify(pq);
    ## Print last k-th largest element (after
    # (inserting last element)
    print(pq[0])
# Driver code
arr = [10, 20, 11, 70, 50, 40, 100, 55]
k = 3
n = len(arr)
kthLargest(arr, n, k)
'''Code is contributed by Rajat Kumar'''

// C# program to find k-th largest element in a 
// stream after every insertion.
using System;
using System.Collections.Generic;
class GFG
{
  static void kthLargest(int[] stream, int n, int k)
  {
 
    // Create a min heap and store first k-1 elements
    // of stream into
    List<int> pq = new List<int>();
 
    // Push first k elements and print "_" (k-1) times
    for (int i = 0; i < k - 1; i++)
    {
      pq.Add(stream[i]);
      Console.Write("_ ");
    }
    pq.Add(stream[k - 1]);
 
    for (int i = k; i < n; i++)
    {
 
      // We must insert last element before we
      // decide last k-th largest output.
      pq.Sort();
      Console.Write(pq[0] + " ");     
      if (stream[i] > pq[0])
      {
        pq.RemoveAt(0);
        pq.Add(stream[i]);
      }  
    }  
 
    // Print last k-th largest element (after
    // (inserting last element)
    pq.Sort();
    Console.Write(pq[0]);
  }
 
  // Driver code
  static void Main()
  {
    int[] arr = {10, 20, 11, 70, 50, 40, 100, 55};
    int k = 3;
    int n = arr.Length;
    kthLargest(arr, n, k);
  }
}
 
// This code is contributed by divyesh072019.

<script>
 
// JavaScript program to find k-th largest element in a
// stream after every insertion.
 
function kthLargest(stream, n, k) {
 
    // Create a min heap and store first k-1 elements
    // of stream into
    let pq = new Array();
 
    // Push first k elements and print "_" (k-1) times
    for (let i = 0; i < k - 1; i++) {
        pq.push(stream[i]);
        document.write("_ ");
    }
    pq.push(stream[k - 1]);
 
    for (let i = k; i < n; i++) {
 
        // We must insert last element before we
        // decide last k-th largest output.
        pq.sort((a, b) => a - b);
        document.write(pq[0] + " ");
        if (stream[i] > pq[0]) {
            pq.shift(0);
            pq.push(stream[i]);
        }
    }
 
    // Print last k-th largest element (after
    // (inserting last element)
    pq.sort((a, b) => a - b);
    document.write(pq[0]);
}
 
// Driver code
 
let arr = [10, 20, 11, 70, 50, 40, 100, 55];
let k = 3;
let n = arr.length;
kthLargest(arr, n, k);
 
// This code is contributed by _saurabh_jaiswal
 
</script>

_ _ 10 11 20 40 50 55

Si el flujo contiene elementos de tipos no primitivos, podemos definir nuestra propia función de compactador y crear una cola de prioridad en consecuencia.