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g1601_1700.s1627_graph_connectivity_with_threshold.readme.md Maven / Gradle / Ivy

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`, 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 directly or indirectly. (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:**

![](https://assets.leetcode.com/uploads/2020/10/09/ex1.jpg)

**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:**

![](https://assets.leetcode.com/uploads/2020/10/10/tmp.jpg)

**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:**

![](https://assets.leetcode.com/uploads/2020/10/17/ex3.jpg)

**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 <= 104
*   `0 <= threshold <= n`
*   1 <= queries.length <= 105
*   `queries[i].length == 2`
*   1 <= ai, bi <= cities
*   ai != bi