# K’th largest element in a stream

• Difficulty Level : Medium
• Last Updated : 21 Jul, 2022

Given an infinite stream of integers, find the k’th largest element at any point of time.
Example:

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

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

Extra space allowed is O(k).

We have discussed different approaches to find k’th largest element in an array in the following posts.
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)
Here we have a stream instead of whole array and we are allowed to store only k elements.

A Simple Solution is to keep an array of size k. The idea is to keep the array sorted so that the k’th largest element can be found in O(1) time (we just need to return first element of array if array is sorted in increasing order)
How to process a new element of stream?
For every new element in stream, check if the new element is smaller than current k’th largest element. If yes, then ignore it. If no, then remove the smallest element from array and insert new element in sorted order. Time complexity of processing a new element is O(k).

A Better Solution is to use a Self Balancing Binary Search Tree of size k. The k’th largest element can be found in O(Logk) time.
How to process a new element of stream?
For every new element in stream, check if the new element is smaller than current k’th largest element. If yes, then ignore it. If no, then remove the smallest element from the tree and insert new element. Time complexity of processing a new element is O(Logk).

An Efficient Solution is to use Min Heap of size k to store k largest elements of stream. The k’th largest element is always at root and can be found in O(1) time.
How to process a new element of stream?
Compare the new element with root of heap. If new element is smaller, then ignore it. Otherwise replace root with new element and call heapify for the root of modified heap. Time complexity of finding the k’th largest element is 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; }`   `    ``// to replace root with new node x and heapify() new` `    ``// root` `    ``void` `replaceMin(``int` `x)` `    ``{` `        ``harr = 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;`   `    ``// If there are more than 1 items, ` `    ``// move the last item to` `    ``// root and call heapify.` `    ``if` `(heap_size > 1) {` `        ``harr = 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 element 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` `}`

## 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 = 4;` `    ``int` `[]arr = { 1, 2, 3, 4, 5, 6 };` `    ``List<``int``> res = getAllKthNumber(arr);` `    ``foreach` `(``int` `x ``in` `res)` `      ``Console.Write(x + ``" "``);` `  ``}` `}`   `// This code is contributed by shikhasingrajput`

Output:

```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```

Implementation using Priority Queue:

## C++

 `// C++ program for the above approach` `#include ` `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 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

 `# Python 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` `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) {` `          ``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;` `    ``}`   `    ``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(nlogn)
Auxiliary Space: O(n),  since n extra space has been taken.