# 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

• 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) {
while (v % i == 0) {
v /= i;
}
}
++i;
}
if (v > 1) {
}
}
}

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) {
}
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
}