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how to determine if the kth largest element of the heap is greater than x

开发者 https://www.devze.com 2023-02-09 15:32 出处:网络
Consider a binary heap containing n numbers (the root stores the greatest number). You are given a positive integer k < n and a number x. You have to determine whether

Consider a binary heap containing n numbers (the root stores the greatest number). You are given a positive integer k < n and a number x. You have to determine whether the kth largest element of the heap is 开发者_高级运维greater than x or not. Your algorithm must take O(k) time. You may use O(k) extra storage


Simple dfs can do the job. We have a counter set to zero. Start from the root and in each iteration check the value of current node; if it is greater than x, then increase the counter and continue the algorithm for one of the child nodes. The algorithm terminates if counter is bigger than equal k or if there is no node left to check. The running time is O(k) because at most k node will be iterated and each iteration is in O(1).

A pseudo-code looks like as follows.

    void CheckNode(Node node,int k, int x, ref int counter)
    {
        if (node.value > x)
        {
            counter++;
            if (counter >= k)
                return;

            CheckNode(node.Left, k, x, ref counter);
            CheckNode(node.Right,k, x, ref counter);
        }
    }

use it:

        counter = 0;
        CheckNode(root,index,val,counter );
        if (counter >= index)
            return true;
        return false;

if node.value < x then all children values are smaller than x and there is no need to check.

As @Eric Mickelsen mentioned in comments worst case running time is exactly 2k-1 (k>0) as follows.

The number of nodes visited with values greater than x will be at most k. Each node visited with value less than x must be a child of a visited node with value greater than x. However, because every node visited except the root must have a parent with value greater than x, the number of nodes of value less than x visited must be at most ((k-1)*2)-(k-1) = k-1, since k-1 of the (k-1)*2 children have values greater than x. This means that we visit k nodes greater than x and k-1 less than x, which is 2k-1.


public class KSmallest2 {

private MinPQ<Integer> minHeap;
private int x;
private int k;
private int count = 0;

public KSmallest2(String filename, int x, int k) {
    this.x = x;
    this.k = k;
    minHeap = new MinPQ<>();
    try {
        Scanner in = new Scanner(new File(filename));
        while (in.hasNext()) {
            minHeap.insert(in.nextInt());
        }
    } catch (FileNotFoundException e) {
        e.printStackTrace();
    }
}

public boolean check(int index) {

    if (index > minHeap.size()) {
        return false;
    }

    if (minHeap.getByIndex(index) < x) {
        count++;
        if (count >= k) {
            return true;
        }

        return  check(2 * index) ||
                check(2 * index + 1);
    }

    return false;
}



public static void main(String[] args) {
    KSmallest2 ks = new KSmallest2("src/main/resources/minheap.txt", 18, 5);
    System.out.println(ks.minHeap);

    System.out.println(ks.check(1));
}

}

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