2.Kth Largest Element in a Stream

Design a class to find the kth largest element in a stream. Note that it is the kth largest element in the sorted order, not the kth distinct element.

Your KthLargest class will have a constructor which accepts an integer k and an integer array nums, which contains initial elements from the stream. For each call to the method KthLargest.add, return the element representing the kth largest element in the stream.

Example:

int k = 3;
int[] arr = [4,5,8,2];
KthLargest kthLargest = new KthLargest(3, arr);
kthLargest.add(3);   // returns 4
kthLargest.add(5);   // returns 5
kthLargest.add(10);  // returns 5
kthLargest.add(9);   // returns 8
kthLargest.add(4);   // returns 8

Solution I: (Using a sorted array of size K)

Time Complexity: Insertion of new element : O(K) Finding the K largest will take O(1) Algo: If the new element is smaller ignore it. Else add in correct position by searching.

class KthLargest
{
public:
    vector<int> v;
    int count;
    KthLargest(int k, vector<int> &nums)
    {
        sort(nums.begin(), nums.end());
        int x = nums.size();

        count = k;
        if (x >= k)
        {
            int i = x - k;
            while (i < nums.size())
            {
                v.push_back(nums[i]);
                i++;
            }
        }
        else
        {
            v = nums;
        }
    }

    int add(int val)
    {
        if (v.size() > 0)
        {
            if (val > v[0] && v.size() == count)
            {
                v[0] = val;
                for (int i = 0; i < v.size() - 1; i++)
                {
                    if (v[i + 1] < v[i])
                    {
                        int temp = v[i];
                        v[i] = v[i + 1];
                        v[i + 1] = temp;
                    }
                }
            }
            else if (v.size() < count)
            {
                v.push_back(val);
                for (int i = 0; i < v.size() - 1; i++)
                {
                    if (v[i + 1] < v[i])
                    {
                        int temp = v[i];
                        v[i] = v[i + 1];
                        v[i + 1] = temp;
                    }
                }
            }
        }
        else
        {
            v.push_back(val);
        }

        return v[0];
    }
};

Solution II: (Using Min Heap)

class KthLargest
{
public:
    priority_queue<int, vector<int>, greater<int>> pq;
    int size;
    
    KthLargest(int k, vector<int> &nums)
    {

        for (int i = 0; i < nums.size(); i++)
        {
            pq.push(nums[i]);
        }

        while (pq.size() > k)
        {
            pq.pop();
        }
        size = k;
    }

    int add(int val)
    {
        pq.push(val);

        if (pq.size() > size)
        {
            pq.pop();
        }

        int res = pq.top();
        return res;
    }
};