# communities problem with union and find

I am trying to solve the following problem:

Input is $$2D$$ array of integers, $$M$$, which corresponds to friendship relations. For example, if $$M[1][2]=1$$, $$1$$ and $$2$$ are friends (assuming symmetry it is also true that $$M[2][1]=1$$). If $$M[2][3]=1$$, then $$\{1,2,3\}$$ is a community. If all other entries are $$0$$, then $$\{0\}$$ is a community by itself, $$\{4\}$$ is a community by itself, etc. One can think of $$M$$ as representation of a graph with $$V$$ nodes, and $$E$$ edges. $$M$$ is $$VxV$$ matrix and the entries with value $$1$$ correspond to edges of an undirected graph. For the above example, the total number of communities is $$V-2$$.

I need to explain the following:

If we solve the above problem with union and find operations (weighted quick union), it will take $$O(E+Vlog(V))$$ time, explain.

I wrote the code for this. I am also including weighted union and find algorithm code as well below for completeness (but it is a standard known algorithm I guess). With the way I implement it, it does not look like this algorithm takes O(E+Vlog(V)) time. I need to understand how can this be done in O(E+Vlog(V)) time.

It is said that find operation takes O(log(V)) time because the tree is almost balanced with weighted quick union algorithm.

This is my code:

The code for weighted quick union and find (given):

public class WeightedQuickUnion {

int[] parent;   // parent[i] = parent of i
int[] size;     // size[i] = number of sites in subtree rooted at i
int count;      // number of components

/**
* Initializes an empty union–find data structure
* Each site is initially in its own
* component.
* n is the number of sites
*/
public WeightedQuickUnion(int n) {
count = n;
parent = new int[n];
size = new int[n];
for (int i = 0; i < n; i++) {
parent[i] = i;
size[i] = 1;
}
}

/**
* Return the number of components.
*
*/
public int count() {
return count;
}

/**
* return the component identifier for the component containing p.
* p the integer representing the item
*/
public int find(int p) {
validate(p);
while (p != parent[p])
p = parent[p];
return p;
}

// validate that p is a valid index
private void validate(int p) {
int n = parent.length;
if (p < 0 || p >= n) {
throw new IllegalArgumentException("index " + p + " is not between 0 and " + (n-1));
}
}

/**
* return true if the the item p and item q are in the same component.
*/
public boolean connected(int p, int q) {
return find(p) == find(q);
}

/**
* Merge the component containing p with the
* the component containing q.
*/
public void union(int p, int q) {
int rootP = find(p);
int rootQ = find(q);

if (rootP == rootQ) return;

// make smaller root point to larger one
if (size[rootP] < size[rootQ]) {
parent[rootP] = rootQ;
size[rootQ] += size[rootP];
}
else {
parent[rootQ] = rootP;
size[rootP] += size[rootQ];
}
count--;
}

• Do you have a question? I'm not clear on what your question is. Can you credit the original source of the question? Debugging your code is outside of the scope of this site; we prefer concise pseudocode that doesn't require knowledge of any particular programming language.
– D.W.
Aug 16 '20 at 21:13