1627. Graph Connectivity With Threshold

Description

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] = [a_i, b_i] if cities a_i and b_i 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 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 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:

2 <= n <= 10⁴
0 <= threshold <= n
1 <= queries.length <= 10⁵
queries[i].length == 2
1 <= a_i, b_i <= cities
a_i != b_i

Solutions

Solution 1: Union-Find

We can enumerate $z$ and its multiples, and use union-find to connect them. In this way, for each query $[a, b]$, we only need to determine whether $a$ and $b$ are in the same connected component.

The time complexity is $O(n \times \log n \times (\alpha(n) + q))$, and the space complexity is $O(n)$. Here, $n$ and $q$ are the number of nodes and queries, respectively, and $\alpha$ is the inverse function of the Ackermann function.

class UnionFind {
    private int[] p;
    private int[] size;

    public UnionFind(int n) {
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            size[i] = 1;
        }
    }

    public int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

    public boolean union(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) {
            return false;
        }
        if (size[pa] > size[pb]) {
            p[pb] = pa;
            size[pa] += size[pb];
        } else {
            p[pa] = pb;
            size[pb] += size[pa];
        }
        return true;
    }
}

class Solution {
    public List<Boolean> areConnected(int n, int threshold, int[][] queries) {
        UnionFind uf = new UnionFind(n + 1);
        for (int a = threshold + 1; a <= n; ++a) {
            for (int b = a + a; b <= n; b += a) {
                uf.union(a, b);
            }
        }
        List<Boolean> ans = new ArrayList<>();
        for (var q : queries) {
            ans.add(uf.find(q[0]) == uf.find(q[1]));
        }
        return ans;
    }
}

class UnionFind {
public:
    UnionFind(int n) {
        p = vector<int>(n);
        size = vector<int>(n, 1);
        iota(p.begin(), p.end(), 0);
    }

    bool unite(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) {
            return false;
        }
        if (size[pa] > size[pb]) {
            p[pb] = pa;
            size[pa] += size[pb];
        } else {
            p[pa] = pb;
            size[pb] += size[pa];
        }
        return true;
    }

    int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

private:
    vector<int> p, size;
};

class Solution {
public:
    vector<bool> areConnected(int n, int threshold, vector<vector<int>>& queries) {
        UnionFind uf(n + 1);
        for (int a = threshold + 1; a <= n; ++a) {
            for (int b = a + a; b <= n; b += a) {
                uf.unite(a, b);
            }
        }
        vector<bool> ans;
        for (auto& q : queries) {
            ans.push_back(uf.find(q[0]) == uf.find(q[1]));
        }
        return ans;
    }
};

class UnionFind:
    def __init__(self, n):
        self.p = list(range(n))
        self.size = [1] * n

    def find(self, x):
        if self.p[x] != x:
            self.p[x] = self.find(self.p[x])
        return self.p[x]

    def union(self, a, b):
        pa, pb = self.find(a), self.find(b)
        if pa == pb:
            return False
        if self.size[pa] > self.size[pb]:
            self.p[pb] = pa
            self.size[pa] += self.size[pb]
        else:
            self.p[pa] = pb
            self.size[pb] += self.size[pa]
        return True


class Solution:
    def areConnected(
        self, n: int, threshold: int, queries: List[List[int]]
    ) -> List[bool]:
        uf = UnionFind(n + 1)
        for a in range(threshold + 1, n + 1):
            for b in range(a + a, n + 1, a):
                uf.union(a, b)
        return [uf.find(a) == uf.find(b) for a, b in queries]

type unionFind struct {
	p, size []int
}

func newUnionFind(n int) *unionFind {
	p := make([]int, n)
	size := make([]int, n)
	for i := range p {
		p[i] = i
		size[i] = 1
	}
	return &unionFind{p, size}
}

func (uf *unionFind) find(x int) int {
	if uf.p[x] != x {
		uf.p[x] = uf.find(uf.p[x])
	}
	return uf.p[x]
}

func (uf *unionFind) union(a, b int) bool {
	pa, pb := uf.find(a), uf.find(b)
	if pa == pb {
		return false
	}
	if uf.size[pa] > uf.size[pb] {
		uf.p[pb] = pa
		uf.size[pa] += uf.size[pb]
	} else {
		uf.p[pa] = pb
		uf.size[pb] += uf.size[pa]
	}
	return true
}

func areConnected(n int, threshold int, queries [][]int) []bool {
	uf := newUnionFind(n + 1)
	for a := threshold + 1; a <= n; a++ {
		for b := a + a; b <= n; b += a {
			uf.union(a, b)
		}
	}
	ans := make([]bool, len(queries))
	for i, q := range queries {
		ans[i] = uf.find(q[0]) == uf.find(q[1])
	}
	return ans
}

class UnionFind {
    p: number[];
    size: number[];
    constructor(n: number) {
        this.p = Array(n)
            .fill(0)
            .map((_, i) => i);
        this.size = Array(n).fill(1);
    }

    find(x: number): number {
        if (this.p[x] !== x) {
            this.p[x] = this.find(this.p[x]);
        }
        return this.p[x];
    }

    union(a: number, b: number): boolean {
        const [pa, pb] = [this.find(a), this.find(b)];
        if (pa === pb) {
            return false;
        }
        if (this.size[pa] > this.size[pb]) {
            this.p[pb] = pa;
            this.size[pa] += this.size[pb];
        } else {
            this.p[pa] = pb;
            this.size[pb] += this.size[pa];
        }
        return true;
    }
}

function areConnected(n: number, threshold: number, queries: number[][]): boolean[] {
    const uf = new UnionFind(n + 1);
    for (let a = threshold + 1; a <= n; ++a) {
        for (let b = a * 2; b <= n; b += a) {
            uf.union(a, b);
        }
    }
    return queries.map(([a, b]) => uf.find(a) === uf.find(b));
}

1627 - Graph Connectivity With Threshold

1627. Graph Connectivity With Threshold

Description

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1627. Graph Connectivity With Threshold

Description

Solutions

All Problems

All Solutions