Graph Connectivity With Threshold

We have n cities labeled from 1 to n. Two different cities with labels x and y are directly connected by a bidirectional road if and only if x and y share a common divisor strictly greater than some threshold. More formally, cities with labels x and y have a road between them if there exists an integer z such that all of the following are true:

Given the two integers, n and threshold, and an array of queries, you must determine for each queries[i] = [ai, bi] if cities ai and bi are connected (i.e. there is some path between them).

Return an array answer, where answer.length == queries.length and answer[i] is true if for the ith query, there is a path between ai and bi, or answer[i] is false if there is no path.

 

Example 1:

Input: n = 6, threshold = 2, queries = [[1,4],[2,5],[3,6]]
Output: [false,false,true]
Explanation: The divisors for each number:
1:   1
2:   1, 2
3:   1, 3
4:   1, 2, 4
5:   1, 5
6:   1, 2, 3, 6
Using the underlined divisors above the threshold, only cities 3 and 6 share a common divisor, so they are the
only ones directly connected. The result of each query:
[1,4]   1 is not connected to 4
[2,5]   2 is not connected to 5
[3,6]   3 is connected to 6 through path 3--6
Example 2:

Input: n = 6, threshold = 0, queries = [[4,5],[3,4],[3,2],[2,6],[1,3]]
Output: [true,true,true,true,true]
Explanation: The divisors for each number are the same as the previous example. However, since the threshold is 0,
all divisors can be used. Since all numbers share 1 as a divisor, all cities are connected.
Example 3:

Input: n = 5, threshold = 1, queries = [[4,5],[4,5],[3,2],[2,3],[3,4]]
Output: [false,false,false,false,false]
Explanation: Only cities 2 and 4 share a common divisor 2 which is strictly greater than the threshold 1, so they are the only ones directly connected.
Please notice that there can be multiple queries for the same pair of nodes [x, y], and that the query [x, y] is equivalent to the query [y, x].
 

Constraints:


Solution:

class Solution {
    static class UF {
        int[] size;
        int[] parent;
        
        public UF(int n) {
            size = new int[n + 1];
            parent = new int[n + 1];
            for (int i = 1; i <= n; i ++) {
                size[i] = 1;
                parent[i] = i;
            }            
        }
        
        public int find(int x) {
            while (parent[x] != x) {
                parent[x] = parent[parent[x]];
                x = parent[x];
            }
            return x;
        }
        
        public void union(int x, int y) {
            int rootX = find(x);
            int rootY = find(y);
            if (rootX == rootY) return;
            if (size[rootX] > size[rootY]) {
                parent[rootY] = rootX;
                size[rootX] += size[rootY];
            } else {
                parent[rootX] = rootY;
                size[rootY] += size[rootX];
            }
            
        }
        
        public boolean connected(int x, int y) {
            return find(x) == find(y);
        }
    }
    
    
    public List<Boolean> areConnected(int n, int threshold, int[][] queries) {
        List<Boolean> result = new ArrayList();
        if (threshold == 0) {
            for (int i = 0; i < queries.length; i ++) {
                result.add(true);
            }
            return result;
        }
        UF cities = new UF(n);
        for (int i = threshold + 1; i <= n; i ++) {
            for (int j = 1; j * i <= n; j ++) {
                if (!cities.connected(i, j * i)) {
                    cities.union(i, j * i);
                }
            }
        }
        for (int[] q : queries) {
            if (cities.connected(q[0], q[1])) {
                result.add(true);
            } else {
                result.add(false);
            }
        }
        return result;
    }
}