703. Kth Largest Element in a Stream

DifficultyEasy

Description

You are part of a university admissions office and need to keep track of the kth highest test score from applicants in real-time. This helps to determine cut-off marks for interviews and admissions dynamically as new applicants submit their scores.

You are tasked to implement a class which, for a given integer k, maintains a stream of test scores and continuously returns the kth highest test score after a new score has been submitted. More specifically, we are looking for the kth highest score in the sorted list of all scores.

Implement the KthLargest class:

KthLargest(int k, int[] nums) Initializes the object with the integer k and the stream of test scores nums.
int add(int val) Adds a new test score val to the stream and returns the element representing the k^th largest element in the pool of test scores so far.

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

Example 2:

Input:
["KthLargest", "add", "add", "add", "add"]
[[4, [7, 7, 7, 7, 8, 3]], [2], [10], [9], [9]]

Output: [null, 7, 7, 7, 8]

Explanation:

KthLargest kthLargest = new KthLargest(4, [7, 7, 7, 7, 8, 3]);
kthLargest.add(2); // return 7
kthLargest.add(10); // return 7
kthLargest.add(9); // return 7
kthLargest.add(9); // return 8

Constraints:

0 <= nums.length <= 10⁴
1 <= k <= nums.length + 1
-10⁴ <= nums[i] <= 10⁴
-10⁴ <= val <= 10⁴
At most 10⁴ calls will be made to add.

Solutions

Solution 1: Priority Queue (Min Heap)

Thinking

We insert into a stream and must report the \(k\)-th largest after each add. Sorting or scanning the whole history on every query is too heavy when both \(n\) and the query count reach \(10^4\).

Only the current top \(k\) values matter; anything smaller can never be the answer. In a size-\(k\) min-heap the top is the smallest of those \(k\), i.e. the \(k\)-th largest overall.

Push each new value and pop if the heap grows past \(k\). Seeding from \(\textit{nums}\) uses the same \(\textit{add}\). Each update is \(O(\log k)\).

We maintain a priority queue (min heap) \(\textit{minQ}\).

Initially, we add the elements of the array \(\textit{nums}\) to \(\textit{minQ}\) one by one, ensuring that the size of \(\textit{minQ}\) does not exceed \(k\). The time complexity is \(O(n \times \log k)\).

Each time a new element is added, if the size of \(\textit{minQ}\) exceeds \(k\), we pop the top element of the heap to ensure that the size of \(\textit{minQ}\) is \(k\). The time complexity is \(O(\log k)\).

In this way, the elements in \(\textit{minQ}\) are the largest \(k\) elements in the array \(\textit{nums}\), and the top element of the heap is the \(k^{th}\) largest element.

The space complexity is \(O(k)\).

Python3JavaC++GoTypeScriptJavaScript

class KthLargest:

    def __init__(self, k: int, nums: List[int]):
        self.k = k
        self.min_q = []
        for x in nums:
            self.add(x)

    def add(self, val: int) -> int:
        heappush(self.min_q, val)
        if len(self.min_q) > self.k:
            heappop(self.min_q)
        return self.min_q[0]


# Your KthLargest object will be instantiated and called as such:
# obj = KthLargest(k, nums)
# param_1 = obj.add(val)

class KthLargest {
    private PriorityQueue<Integer> minQ;
    private int k;

    public KthLargest(int k, int[] nums) {
        this.k = k;
        minQ = new PriorityQueue<>(k);
        for (int x : nums) {
            add(x);
        }
    }

    public int add(int val) {
        minQ.offer(val);
        if (minQ.size() > k) {
            minQ.poll();
        }
        return minQ.peek();
    }
}

/**
 * Your KthLargest object will be instantiated and called as such:
 * KthLargest obj = new KthLargest(k, nums);
 * int param_1 = obj.add(val);
 */

class KthLargest {
public:
    KthLargest(int k, vector<int>& nums) {
        this->k = k;
        for (int x : nums) {
            add(x);
        }
    }

    int add(int val) {
        minQ.push(val);
        if (minQ.size() > k) {
            minQ.pop();
        }
        return minQ.top();
    }

private:
    int k;
    priority_queue<int, vector<int>, greater<int>> minQ;
};

/**
 * Your KthLargest object will be instantiated and called as such:
 * KthLargest* obj = new KthLargest(k, nums);
 * int param_1 = obj->add(val);
 */

type KthLargest struct {
    k    int
    minQ hp
}

func Constructor(k int, nums []int) KthLargest {
    minQ := hp{}
    this := KthLargest{k, minQ}
    for _, x := range nums {
        this.Add(x)
    }
    return this
}

func (this *KthLargest) Add(val int) int {
    heap.Push(&this.minQ, val)
    if this.minQ.Len() > this.k {
        heap.Pop(&this.minQ)
    }
    return this.minQ.IntSlice[0]
}

type hp struct{ sort.IntSlice }

func (h *hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
func (h *hp) Pop() interface{} {
    old := h.IntSlice
    n := len(old)
    x := old[n-1]
    h.IntSlice = old[0 : n-1]
    return x
}
func (h *hp) Push(x interface{}) {
    h.IntSlice = append(h.IntSlice, x.(int))
}

/**
 * Your KthLargest object will be instantiated and called as such:
 * obj := Constructor(k, nums);
 * param_1 := obj.Add(val);
 */

class KthLargest {
    #k: number = 0;
    #minQ = new MinPriorityQueue<number>();

    constructor(k: number, nums: number[]) {
        this.#k = k;
        for (const x of nums) {
            this.add(x);
        }
    }

    add(val: number): number {
        this.#minQ.enqueue(val);
        if (this.#minQ.size() > this.#k) {
            this.#minQ.dequeue();
        }
        return this.#minQ.front();
    }
}

/**
 * Your KthLargest object will be instantiated and called as such:
 * var obj = new KthLargest(k, nums)
 * var param_1 = obj.add(val)
 */

/**
 * @param {number} k
 * @param {number[]} nums
 */
var KthLargest = function (k, nums) {
    this.k = k;
    this.minQ = new MinPriorityQueue();
    for (const x of nums) {
        this.add(x);
    }
};

/**
 * @param {number} val
 * @return {number}
 */
KthLargest.prototype.add = function (val) {
    this.minQ.enqueue(val);
    if (this.minQ.size() > this.k) {
        this.minQ.dequeue();
    }
    return this.minQ.front();
};

/**
 * Your KthLargest object will be instantiated and called as such:
 * var obj = new KthLargest(k, nums)
 * var param_1 = obj.add(val)
 */

703. Kth Largest Element in a Stream

Description

Solutions

Solution 1: Priority Queue (Min Heap)

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