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K’th largest element in a stream

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  • Difficulty Level : Medium
  • Last Updated : 12 Oct, 2022
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Given an infinite stream of integers, find the Kth largest element at any point of time.

Note: Here we have a stream instead of a whole array and we are allowed to store only K elements.

Examples: 

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

Input: stream[] = {2, 5, 1, 7, 9, . . .}, K = 2
Output: {_, 2, 2, 5, 7, . . .} 

Naive Approach: To solve the problem follow the below idea:

Keep an array of size K. The idea is to keep the array sorted so that the Kth largest element can be found in O(1) time (we just need to return the first element of the array, if the array is sorted in increasing order

How to process a new element of the stream? 

For every new element in the stream, check if the new element is smaller than the current Kth largest element. If yes, then ignore it. If no, then remove the smallest element from the array and insert the new element in sorted order. 

The time complexity of processing a new element is O(K)

Kth largest element in a stream using a self-balancing binary search tree:

To solve the problem follow the below idea:

Create a self-balancing binary search tree and for every new element in the stream, check if the new element is smaller than the current k’th largest element. If yes, then ignore it. If no, then remove the smallest element from the tree and insert a new element. 

The Kth largest element can be found in O(log K) time.

Kth largest element in a stream using a Min-Heap:

To solve the problem follow the below idea:

An Efficient Solution is to use a Min Heap of size K to store K largest elements of the stream. The Kth largest element is always at the root and can be found in O(1) time

How to process a new element of the stream? 

Compare the new element with the root of the heap. If a new element is smaller, then ignore it. Otherwise, replace the root with a new element and call heapify for the root of the modified heap

Below is the implementation of the above approach:

CPP




// A C++ program to find k'th
// smallest element in a stream
#include <bits/stdc++.h>
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 value.
    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, vector<int>& A)
{
    // 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);
 
    for (auto& x : A) {
 
        // Nothing much to do for first k-1 elements
        if (count < k - 1) {
            arr[count] = x;
            count++;
 
            cout << "Kth largest element is -1 " << endl;
        }
 
        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 << "Kth largest element is "
                 << mh.getMin() << endl;
            count++;
        }
    }
}
 
// Driver code
int main()
{
    vector<int> arr = { 1, 2, 3, 4, 5, 6 };
    int K = 3;
 
    // Function call
    kthLargest(K, arr);
    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 element of heap
    itself the kth largest largest element
 
    */
    static PriorityQueue<Integer> min;
    static int k;
 
    static List<Integer> getAllKthNumber(int arr[])
    {
 
        // list to store kth largest number
        List<Integer> 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 = 3;
 
        int arr[] = { 1, 2, 3, 4, 5, 6 };
 
        // Function call
        List<Integer> res = getAllKthNumber(arr);
 
        for (int x : res)
            System.out.println("Kth largest element is "
                               + x);
    }
 
    // This code is Contributed by Pradeep Mondal P
}

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG {
 
    /*
    using min heap DS
 
    how data are stored in min Heap DS
           1
         2   3
    if k==3 , then top element of heap
    itself the kth largest largest element
 
    */
    static Queue<int> min;
    static int k;
 
    static List<int> getAllKthNumber(int[] arr)
    {
 
        // list to store kth largest number
        List<int> list = new List<int>();
 
        // one by one adding values to the min heap
        foreach(int val in arr)
        {
 
            // if the heap size is less than k , we add to
            // the heap
            if (min.Count < k)
                min.Enqueue(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.Dequeue();
                    min.Enqueue(val);
                }
            }
 
            // if heap size >=k we add
            // kth largest element
            // otherwise -1
 
            if (min.Count >= k)
                list.Add(min.Peek());
            else
                list.Add(-1);
        }
        return list;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        min = new Queue<int>();
        k = 3;
        int[] arr = { 1, 2, 3, 4, 5, 6 };
 
        // Function call
        List<int> res = getAllKthNumber(arr);
        foreach(int x in res) Console.Write(
            "Kth largest element is " + x + "\n");
    }
}
 
// This code is contributed by shikhasingrajput

Output

Kth largest element is -1 
Kth largest element is -1 
Kth largest element is 1
Kth largest element is 2
Kth largest element is 3
Kth largest element is 4

Time Complexity: O(N * log K)
Auxiliary Space: O(K)

Below is the implementation of the above approach using priority-queue:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
vector<int> kthLargest(int k, int arr[], int n)
{
    vector<int> ans(n);
 
    // Creating a min-heap using priority queue
    priority_queue<int, vector<int>, greater<int> > 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<int> v = kthLargest(k, arr, n);
    for (auto it : v)
        cout << it << " ";
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG {
 
    static int[] kthLargest(int k, int arr[], int n)
    {
        int[] ans = new int[n];
 
        // Creating a min-heap using priority queue
        PriorityQueue<Integer> pq
            = new PriorityQueue<>((a, b) -> a - b);
 
        // 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.add(arr[i]);
            else {
                if (arr[i] > pq.peek()) {
                    pq.remove();
                    pq.add(arr[i]);
                }
            }
 
            // If size is less than k
            if (pq.size() < k)
                ans[i] = -1;
            else
                ans[i] = pq.peek();
        }
 
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 6;
        int arr[] = { 1, 2, 3, 4, 5, 6 };
        int k = 4;
 
        // Function call
        int[] v = kthLargest(k, arr, n);
        for (int it : v)
            System.out.print(it + " ");
    }
}
 
// This code is contributed by shikhasingrajput

Python3




# Python3 program for the above approach
from queue import PriorityQueue
 
 
def kthLargest(k,  arr,  n):
    ans = [0]*n
 
    # Creating a min-heap using priority queue
    pq = PriorityQueue()
 
    # Iterating through each element
    for i in range(n):
        # If size of priority
        # queue is less than k
        if (pq.qsize() < k):
            pq.put(arr[i])
        else:
            if (arr[i] > pq.queue[0]):
                pq.get()
                pq.put(arr[i])
 
    # If size is less than k
        if (pq.qsize() < k):
            ans[i] = -1
        else:
            ans[i] = pq.queue[0]
 
    return ans
 
 
# Driver Code
if __name__ == "__main__":
    n = 6
    arr = [1, 2, 3, 4, 5, 6]
    k = 4
 
    # Function call
    v = kthLargest(k, arr, n)
    print(*v)
 
# This code is contributed by Lovely Jain

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG {
 
    static int[] kthLargest(int k, int[] arr, int n)
    {
        int[] ans = new int[n];
 
        // Creating a min-heap using priority queue
        List<int> pq = new List<int>();
 
        // Iterating through each element
        for (int i = 0; i < n; i++) {
 
            // If size of priority
            // queue is less than k
            if (pq.Count < k)
                pq.Add(arr[i]);
            else {
                if (arr[i] > pq[0]) {
                    pq.Sort();
                    pq.RemoveAt(0);
                    pq.Add(arr[i]);
                }
            }
 
            // If size is less than k
            if (pq.Count < k)
                ans[i] = -1;
            else
                ans[i] = pq[0];
        }
 
        return ans;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int n = 6;
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int k = 4;
 
        // Function call
        int[] v = kthLargest(k, arr, n);
        foreach(int it in v) Console.Write(it + " ");
    }
}
 
// This code contributed by shikhasingrajput

Output

-1 -1 -1 1 2 3 

Time Complexity: O(N * log K) 
Auxiliary Space: O(K)

Related Articles:
K’th Smallest/Largest Element in Unsorted Array | Set 1 
K’th Smallest/Largest Element in Unsorted Array | Set 2 (Expected Linear Time)
K’th Smallest/Largest Element in Unsorted Array | Set 3 (Worst Case Linear Time)

This article is contributed by Shivam Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


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