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2709. Greatest Common Divisor Traversal

Description

You are given a 0-indexed integer array nums, and you are allowed to traverse between its indices. You can traverse between index i and index j, i != j, if and only if gcd(nums[i], nums[j]) > 1, where gcd is the greatest common divisor.

Your task is to determine if for every pair of indices i and j in nums, where i < j, there exists a sequence of traversals that can take us from i to j.

Return true if it is possible to traverse between all such pairs of indices, or false otherwise.

 

Example 1:

Input: nums = [2,3,6]
Output: true
Explanation: In this example, there are 3 possible pairs of indices: (0, 1), (0, 2), and (1, 2).
To go from index 0 to index 1, we can use the sequence of traversals 0 -> 2 -> 1, where we move from index 0 to index 2 because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 > 1, and then move from index 2 to index 1 because gcd(nums[2], nums[1]) = gcd(6, 3) = 3 > 1.
To go from index 0 to index 2, we can just go directly because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 > 1. Likewise, to go from index 1 to index 2, we can just go directly because gcd(nums[1], nums[2]) = gcd(3, 6) = 3 > 1.

Example 2:

Input: nums = [3,9,5]
Output: false
Explanation: No sequence of traversals can take us from index 0 to index 2 in this example. So, we return false.

Example 3:

Input: nums = [4,3,12,8]
Output: true
Explanation: There are 6 possible pairs of indices to traverse between: (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), and (2, 3). A valid sequence of traversals exists for each pair, so we return true.

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 105

Solutions

Solution 1

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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


mx = 100010
p = defaultdict(list)
for x in range(1, mx + 1):
    v = x
    i = 2
    while i <= v // i:
        if v % i == 0:
            p[x].append(i)
            while v % i == 0:
                v //= i
        i += 1
    if v > 1:
        p[x].append(v)


class Solution:
    def canTraverseAllPairs(self, nums: List[int]) -> bool:
        n = len(nums)
        m = max(nums)
        uf = UnionFind(n + m + 1)
        for i, x in enumerate(nums):
            for j in p[x]:
                uf.union(i, j + n)
        return len(set(uf.find(i) for i in range(n))) == 1

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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 {
    private static final int MX = 100010;
    private static final List<Integer>[] P = new List[MX];

    static {
        Arrays.setAll(P, k -> new ArrayList<>());
        for (int x = 1; x < MX; ++x) {
            int v = x;
            int i = 2;
            while (i <= v / i) {
                if (v % i == 0) {
                    P[x].add(i);
                    while (v % i == 0) {
                        v /= i;
                    }
                }
                ++i;
            }
            if (v > 1) {
                P[x].add(v);
            }
        }
    }

    public boolean canTraverseAllPairs(int[] nums) {
        int m = Arrays.stream(nums).max().getAsInt();
        int n = nums.length;
        UnionFind uf = new UnionFind(n + m + 1);
        for (int i = 0; i < n; ++i) {
            for (int j : P[nums[i]]) {
                uf.union(i, j + n);
            }
        }
        Set<Integer> s = new HashSet<>();
        for (int i = 0; i < n; ++i) {
            s.add(uf.find(i));
        }
        return s.size() == 1;
    }
}

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int MX = 100010;
vector<int> P[100010];

int init = []() {
    for (int x = 1; x < MX; ++x) {
        int v = x;
        int i = 2;
        while (i <= v / i) {
            if (v % i == 0) {
                P[x].push_back(i);
                while (v % i == 0) {
                    v /= i;
                }
            }
            ++i;
        }
        if (v > 1) {
            P[x].push_back(v);
        }
    }
    return 0;
}();

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:
    bool canTraverseAllPairs(vector<int>& nums) {
        int m = *max_element(nums.begin(), nums.end());
        int n = nums.size();
        UnionFind uf(m + n + 1);
        for (int i = 0; i < n; ++i) {
            for (int j : P[nums[i]]) {
                uf.unite(i, j + n);
            }
        }
        unordered_set<int> s;
        for (int i = 0; i < n; ++i) {
            s.insert(uf.find(i));
        }
        return s.size() == 1;
    }
};

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const mx = 100010

var p = make([][]int, mx)

func init() {
    for x := 1; x < mx; x++ {
        v := x
        i := 2
        for i <= v/i {
            if v%i == 0 {
                p[x] = append(p[x], i)
                for v%i == 0 {
                    v /= i
                }
            }
            i++
        }
        if v > 1 {
            p[x] = append(p[x], v)
        }
    }
}

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 canTraverseAllPairs(nums []int) bool {
    m := slices.Max(nums)
    n := len(nums)
    uf := newUnionFind(m + n + 1)
    for i, x := range nums {
        for _, j := range p[x] {
            uf.union(i, j+n)
        }
    }
    s := map[int]bool{}
    for i := 0; i < n; i++ {
        s[uf.find(i)] = true
    }
    return len(s) == 1
}

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