# Graph Connectivity With Threshold

This page explains Java solution to problem `Graph Connectivity With Threshold` using `Union And Find` algorithm.

## Problem Statement

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:

• `x % z == 0`,
• `y % z == 0`, and
• `z > threshold`.

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]

Example 3:

Input: n = 5, threshold = 1, queries = [[4,5],[4,5],[3,2],[2,3],[3,4]]
Output: [false,false,false,false,false]

## Solution

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``````
package com.vc.hard;

import java.util.ArrayList;
import java.util.List;

class GraphConnectivityWithThreshold {
public List<Boolean> areConnected(int n, int threshold, int[][] queries) {
if(queries == null) return new ArrayList<>();

int[] parent = new int[n + 1];
for(int i = 0; i <= n; i++) parent[i] = i;

for(int i = threshold + 1; i <= n; i++) {
for(int j = i; j <= n; j+=i) {
int iParent = find(i, parent);
int jParent = find(j, parent);
parent[iParent] = jParent;
}
}

List<Boolean> res = new ArrayList<>();
for(int[] query: queries) {
int xParent = find(query, parent);
int yParent = find(query, parent);
}
return res;
}

private int find(int x, int[] parent) {
if(x == parent[x]) return x;
else {
parent[x] = find(parent[x], parent);
return parent[x];
}
}
}
``````

## Time Complexity

O(N log N + Q) Where
N is total number of cities
Q is total number of queries

## Space Complexity

O(N + Q) Where
N is total number of cities
Q is total number of queries