C++ Program to Find the K’th largest element in a stream

Last Updated : 25 Oct, 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 <climits>
#include <iostream>
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)
{
    // 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;
}

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

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

Implementation using Priority Queue:

CPP

// 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;
}

Output

-1 -1 -1 1 2 3

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

Please refer complete article on K’th largest element in a stream for more details!

How to Find the Last Element in a Map in C++?

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