PRINCETON Course Algorithms Interview Questions: Union Find

可以看下Coursera PRINCETON Algorithm Part1课程,
或者,Algorithms一书。
网盘分享:[Course Vedio & Slides] [pdf]

Coursera上不仅有课后作业,还有面试题目,下面分析一下所给题目。

首先给出一份带有路径压缩的并查集代码。

const int N = 111111;

int id[N];

void Init()
{
    for (int i = 0; i < N; ++i) {
        id[i] = i;
    }
}

int Find(int p)
{
    if (id[p] == p) return id[p];
    else {
        return id[p] = Find(id[p]); // compress the path
    }
}

bool Connected(int p, int q)
{
    return Find(p) == Find(q);
}

void Union(int p, int q)
{
    int i = Find(p);
    int j = Find(q);
    id[i] = j;
}

Reference

Coursera Algorithms Part1 Union-Find Interview Question
https://class.coursera.org/algs4partI-009/quiz/attempt?quiz_id=89

Question 1

Social network connectivity. Given a social network containing N members and a log file containing M timestamps at which times pairs of members formed friendships, design an algorithm to determine the earliest time at which all members are connected (i.e., every member is a friend of a friend of a friend … of a friend). Assume that the log file is sorted by timestamp and that friendship is an equivalence relation. The running time of your algorithm should be MlogN or better and use extra space proportional to N.

Solution

初始森林个数为N,log file 的每一行输入t u v,做union(u, v)。而在union(u, v)之前,检查connected(u, v), if false, then N -= 1。当N减为1时,则当前t即为the earliest time
使用路径压缩的并查集,时间复杂度可以达到O(MlogN),而内存也满足O(N)

Question 2

Union-find with specific canonical element. Add a method find() to the union-find data type so that find(i) returns the largest element in the connected component containing i. The operations, union(), connected(), and find() should all take logarithmic time or better.
For example, if one of the connected components is {1,2,6,9}, then the find() method should return 9 for each of the four elements in the connected components because 9 is larger 1, 2, and 6.

Solution

保证每一个树(集合)的根节点值最大,所以在union的时候将根节点值大的作为新树的根节点。find只需要找到根节点的值。connected函数还是和原来的一样。
使用路径压缩,复杂度为log级别。

Question 3

Successor with delete. Given a set of N integers S={0,1,…,N−1} and a sequence of requests of the following form:

  • Remove x from S
  • Find the successor of x: the smallest y in S such that y≥x.
    design a data type so that all operations (except construction) should take logarithmic time or better.

Solution

可以使用set之类的平衡二叉树数据结构,很容易实现上述操作。
如果是使用Union-Find算法的话…好吧,最近撸论文demo实现,脑子坏了,想了一会没想出,网上找了一下。看下面的图就很直观了。

Q3-Illustration

find只需返回根节点的值。

Question 4

Union-by-size. Develop a union-find implementation that uses the same basic strategy as weighted quick-union but keeps track of tree height and always links the shorter tree to the taller one. Prove a lgN upper bound on the height of the trees for N sites with your algorithm.

Solution

在每一个节点增加一个数据域,height,记录其高度。每次做union操作时,更新根节点的height值。
时间复杂度证明:对于N个节点,按照每次将高度较大的作为根节点,最坏情况下,每次合并的两棵树高度相同,然后新树的高度+1。所以,有树最高lgN,即时间复杂最坏为lgN

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