Kth Largest Element in a Stream - Min Heap

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Problem Statement:

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.

Implement KthLargest class:

• KthLargest(int k, int[] nums) Initializes the object with the integer k and the stream of integers nums.
• int add(int val) Appends the integer val to the stream and returns the element representing the kth largest element in the stream.

 

Example 1:

Input
["KthLargest", "add", "add", "add", "add", "add"]
[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]
Output
[null, 4, 5, 5, 8, 8]

Explanation
KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
kthLargest.add(3);   // return 4
kthLargest.add(5);   // return 5
kthLargest.add(10);  // return 5
kthLargest.add(9);   // return 8
kthLargest.add(4);   // return 8

 

Constraints:

• 1 <= k <= 104
• 0 <= nums.length <= 104
• -104 <= nums[i] <= 104
• -104 <= val <= 104
• At most 104 calls will be made to add.
• It is guaranteed that there will be at least k elements in the array when you search for the kth element.

Solution:

Maintain the highest K elements in a min Heap. Then the root node is always the answer.

class KthLargest {
public:
    priority_queue<int,vector<int>,greater<int>> pq;
    int k;
    KthLargest(int k, vector<int>& nums): k(k)
    {
        for (int n: nums) pq.push(n);
        while (pq.size()>k) pq.pop();
    }
    
    int add(int val) 
    {
        pq.push(val);
        while(pq.size()>k) pq.pop();
        return pq.top();
    }
};