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Formatted question description: https://leetcode.ca/all/1627.html
1627. Graph Connectivity With Threshold (Hard)
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
, andz > threshold
.
Given the two integers, n
and threshold
, and an array of queries
, you must determine for each queries[i] = [a_{i}, b_{i}]
if cities a_{i}
and b_{i}
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 i^{th}
query, there is a path between a_{i}
and b_{i}
, 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 36
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:
2 <= n <= 10^{4}
0 <= threshold <= n
1 <= queries.length <= 10^{5}
queries[i].length == 2
1 <= a_{i}, b_{i} <= cities
a_{i} != b_{i}
Related Topics:
Math, Union Find
Solution 1. Union Find
Intuition:
 If two numbers share the same factor, use union find to connect them.
 For each query, use union find to check if they are connected.
For #1, the brute force way is to check all combination pairs of cities which is O(N^2)
time complexity and will get TLE.
A more efficient way is to use similar idea as in Sieve of Eratosthenes.
For each number i
in [1 + threshold, n]
, we connect i
with multiples of i
(i.e. 2 * i, 3 * i, 4 * i, ...
).
In the worst case where threshold = 0
, iterating all the city pairs cost N + N / 2 + N / 3 + ... + 1 = N * (1 + 1 / 2 + 1 / 3 + ... + 1 / N)
. 1 + 1 / 2 + 1 / 3 + ... + 1 / N
is a harmonic series and bounded by logN
. So the iteration takes O(NlogN)
.
// OJ: https://leetcode.com/problems/graphconnectivitywiththreshold/
// Time: O(NlogN + Q)
// Space: O(N)
// Ref: https://leetcode.com/problems/graphconnectivitywiththreshold/discuss/899595/C%2B%2BJavaPythonUnionFindO(N*logN%2Bq)
class UnionFind {
vector<int> id;
public:
UnionFind(int n) : id(n) {
iota(begin(id), end(id), 0);
}
int find(int x) {
return id[x] == x ? x : (id[x] = find(id[x]));
}
void connect(int a, int b) {
int p = find(a), q = find(b);
if (p == q) return;
id[p] = q;
}
bool connected(int a, int b) {
return find(a) == find(b);
}
};
class Solution {
public:
vector<bool> areConnected(int n, int threshold, vector<vector<int>>& Q) {
unordered_map<int, int> m;
UnionFind uf(n);
for (int i = 1 + threshold; i <= n; ++i) {
for (int j = 2 * i; j <= n; j += i) {
uf.connect(i  1, j  1);
}
}
vector<bool> ans;
for (auto &q : Q) ans.push_back(uf.connected(q[0]  1, q[1]  1));
return ans;
}
};

class Solution { public List<Boolean> areConnected(int n, int threshold, int[][] queries) { int[] parents = new int[n + 1]; for (int i = 1; i <= n; i++) parents[i] = i; int max = n / 2; for (int i = threshold + 1; i <= max; i++) { for (int j = i * 2; j <= n; j += i) union(parents, i, j); } List<Boolean> list = new ArrayList<Boolean>(); for (int[] query : queries) list.add(find(parents, query[0]) == find(parents, query[1])); return list; } public void union(int[] parents, int index1, int index2) { parents[find(parents, index1)] = find(parents, index2); } public int find(int[] parents, int index) { while (parents[index] != index) { parents[index] = find(parents, parents[index]); index = parents[index]; } return index; } }

// OJ: https://leetcode.com/problems/graphconnectivitywiththreshold/ // Time: O(NlogN + Q) // Space: O(N) // Ref: https://leetcode.com/problems/graphconnectivitywiththreshold/discuss/899595 class UnionFind { vector<int> id; public: UnionFind(int n) : id(n) { iota(begin(id), end(id), 0); } int find(int x) { return id[x] == x ? x : (id[x] = find(id[x])); } void connect(int a, int b) { id[find(a)] = find(b); } bool connected(int a, int b) { return find(a) == find(b); } }; class Solution { public: vector<bool> areConnected(int n, int threshold, vector<vector<int>>& Q) { UnionFind uf(n + 1); for (int d = threshold + 1; d <= n / 2; ++d) { // Enumerate all possible common divisors for (int i = 2 * d; i <= n; i += d) { // Union all the numbers sharing the same divisors together uf.connect(d, i); } } vector<bool> ans; for (auto &q : Q) ans.push_back(uf.connected(q[0], q[1])); return ans; } };