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--;
}
