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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 <= 1051 <= nums[i] <= 105
Solutions
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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; } } -
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; } }; -
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 -
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 }