Kth Largest Numbers From Stream - Amazon Top Interview Questions


Problem Statement :


Implement a data structure with the following methods:

KthLargestStream(int[] nums, int k) which constructs the instance.
int add(int val) which adds val to nums and returns the k (0-indexed) largest value in nums
Constraints

1 ≤ n ≤ 100,000 where n is the length of nums
0 ≤ m ≤ 100,000 where m is the number of calls to add
k + 1 ≤ n

Example 1

Input

methods = ["constructor", "add", "add", "add", "add"]

arguments = [[[1, 2, 4, 3], 3], [5], [6], [7], [8]]`

Output

[None, 2, 3, 4, 5]

Explanation

s = KthLargestStream([1,2,4,3],3)

s.add(5) == 2 # Stream has [1, 2, 3, 4, 5]

s.add(6) == 3 # Stream has [1, 2, 3, 4, 5, 6]

s.add(7) == 4 # Stream has [1, 2, 3, 4, 5, 6, 7]

s.add(8) == 5 # Stream has [1, 2, 3, 4, 5, 6, 7, 8]



Solution :



title-img




                        Solution in C++ :

class KthLargestStream {
    public:
    KthLargestStream(vector<int>& nums, int k) {
        K = k + 1;
        for (auto& num : nums) {
            pq.push(num);
            if (pq.size() > K) {
                pq.pop();
            }
        }
    }

    int add(int val) {
        pq.push(val);
        if (pq.size() > K) {
            pq.pop();
        }

        return pq.top();
    }

    private:
    priority_queue<int, vector<int>, greater<int>> pq;
    int K;
};
                    


                        Solution in Java :

import java.util.*;

class KthLargestStream {
    PriorityQueue<Integer> pq;
    int k;
    public KthLargestStream(int[] nums, int k) {
        this.pq = new PriorityQueue<Integer>();
        this.k = k;
        for (int num : nums) {
            pq.add(num);
            if (pq.size() > k + 1)
                pq.poll();
        }
    }

    public int add(int val) {
        pq.add(val);
        if (pq.size() > k + 1)
            pq.poll();
        return pq.peek();
    }
}
                    


                        Solution in Python : 
                            
class KthLargestStream:
    def __init__(self, nums, k):
        heapq.heapify(nums)

        while len(nums) > k + 1:
            heapq.heappop(nums)

        self.nums = nums

    def add(self, val):
        heapq.heappushpop(self.nums, val)

        return self.nums[0]
                    


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