Greatest Common Divisor Traversal
HardUpdated: Aug 2, 2025
Practice on:
Problem
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,_orfalse otherwise.
Examples
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 <= 10^51 <= nums[i] <= 10^5
Solution
Method 1 – Union-Find with Prime Factorization
Intuition
Two indices are connected if their numbers share a common prime factor. If all indices are in the same connected component (via shared prime factors), then traversal is possible between any pair. We can use Union-Find (DSU) to group indices by their prime factors.
Approach
- For each number in
nums, factorize it into its prime factors. - For each prime factor, union the index of the number and the prime factor in DSU.
- After processing all numbers, check if all indices are in the same connected component by comparing their roots in DSU.
Code
C++
class Solution {
public:
bool canTraverseAllPairs(vector<int>& nums) {
int n = nums.size();
if (n == 1) return true;
unordered_map<int, int> prime_to_idx;
vector<int> p(n);
iota(p.begin(), p.end(), 0);
function<int(int)> find = [&](int x) { return p[x] == x ? x : p[x] = find(p[x]); };
auto factor = [](int x) {
vector<int> res;
for (int d = 2; d * d <= x; ++d) {
if (x % d == 0) {
res.push_back(d);
while (x % d == 0) x /= d;
}
}
if (x > 1) res.push_back(x);
return res;
};
for (int i = 0; i < n; ++i) {
for (int pf : factor(nums[i])) {
if (prime_to_idx.count(pf)) {
p[find(i)] = find(prime_to_idx[pf]);
} else {
prime_to_idx[pf] = i;
}
}
}
int root = find(0);
for (int i = 1; i < n; ++i) {
if (find(i) != root) return false;
}
return true;
}
};
Go
func canTraverseAllPairs(nums []int) bool {
n := len(nums)
if n == 1 {
return true
}
p := make([]int, n)
for i := range p { p[i] = i }
var find func(int) int
find = func(x int) int {
if p[x] != x { p[x] = find(p[x]) }
return p[x]
}
primeToIdx := map[int]int{}
factor := func(x int) []int {
res := []int{}
for d := 2; d*d <= x; d++ {
if x%d == 0 {
res = append(res, d)
for x%d == 0 { x /= d }
}
}
if x > 1 { res = append(res, x) }
return res
}
for i, v := range nums {
for _, pf := range factor(v) {
if idx, ok := primeToIdx[pf]; ok {
p[find(i)] = find(idx)
} else {
primeToIdx[pf] = i
}
}
}
root := find(0)
for i := 1; i < n; i++ {
if find(i) != root {
return false
}
}
return true
}
Java
class Solution {
public boolean canTraverseAllPairs(int[] nums) {
int n = nums.length;
if (n == 1) return true;
int[] p = new int[n];
for (int i = 0; i < n; i++) p[i] = i;
java.util.Map<Integer, Integer> primeToIdx = new java.util.HashMap<>();
java.util.function.IntUnaryOperator find = new java.util.function.IntUnaryOperator() {
public int applyAsInt(int x) { return p[x] == x ? x : (p[x] = applyAsInt(p[x])); }
};
java.util.function.Function<Integer, java.util.List<Integer>> factor = x -> {
java.util.List<Integer> res = new java.util.ArrayList<>();
for (int d = 2; d * d <= x; d++) {
if (x % d == 0) {
res.add(d);
while (x % d == 0) x /= d;
}
}
if (x > 1) res.add(x);
return res;
};
for (int i = 0; i < n; i++) {
for (int pf : factor.apply(nums[i])) {
if (primeToIdx.containsKey(pf)) {
p[find.applyAsInt(i)] = find.applyAsInt(primeToIdx.get(pf));
} else {
primeToIdx.put(pf, i);
}
}
}
int root = find.applyAsInt(0);
for (int i = 1; i < n; i++) {
if (find.applyAsInt(i) != root) return false;
}
return true;
}
}
Kotlin
class Solution {
fun canTraverseAllPairs(nums: IntArray): Boolean {
val n = nums.size
if (n == 1) return true
val p = IntArray(n) { it }
fun find(x: Int): Int = if (p[x] == x) x else { p[x] = find(p[x]); p[x] }
val primeToIdx = mutableMapOf<Int, Int>()
fun factor(x: Int): List<Int> {
var x = x
val res = mutableListOf<Int>()
var d = 2
while (d * d <= x) {
if (x % d == 0) {
res.add(d)
while (x % d == 0) x /= d
}
d++
}
if (x > 1) res.add(x)
return res
}
for (i in 0 until n) {
for (pf in factor(nums[i])) {
if (primeToIdx.containsKey(pf)) {
p[find(i)] = find(primeToIdx[pf]!!)
} else {
primeToIdx[pf] = i
}
}
}
val root = find(0)
for (i in 1 until n) {
if (find(i) != root) return false
}
return true
}
}
Python
class Solution:
def canTraverseAllPairs(self, nums: list[int]) -> bool:
n = len(nums)
if n == 1:
return True
p = list(range(n))
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]
from collections import defaultdict
prime_to_idx = dict()
def factor(x):
res = set()
d = 2
while d * d <= x:
if x % d == 0:
res.add(d)
while x % d == 0:
x //= d
d += 1
if x > 1:
res.add(x)
return res
for i, v in enumerate(nums):
for pf in factor(v):
if pf in prime_to_idx:
p[find(i)] = find(prime_to_idx[pf])
else:
prime_to_idx[pf] = i
root = find(0)
for i in range(1, n):
if find(i) != root:
return False
return True
Rust
impl Solution {
pub fn can_traverse_all_pairs(nums: Vec<i32>) -> bool {
let n = nums.len();
if n == 1 { return true; }
let mut p: Vec<usize> = (0..n).collect();
fn find(p: &mut Vec<usize>, x: usize) -> usize {
if p[x] != x { p[x] = find(p, p[x]); }
p[x]
}
use std::collections::HashMap;
let mut prime_to_idx = HashMap::new();
fn factor(mut x: i32) -> Vec<i32> {
let mut res = vec![];
let mut d = 2;
while d * d <= x {
if x % d == 0 {
res.push(d);
while x % d == 0 { x /= d; }
}
d += 1;
}
if x > 1 { res.push(x); }
res
}
for (i, &v) in nums.iter().enumerate() {
for pf in factor(v) {
if let Some(&idx) = prime_to_idx.get(&pf) {
let ri = find(&mut p, i);
let rj = find(&mut p, idx);
p[ri] = rj;
} else {
prime_to_idx.insert(pf, i);
}
}
}
let root = find(&mut p, 0);
for i in 1..n {
if find(&mut p, i) != root {
return false;
}
}
true
}
}
TypeScript
class Solution {
canTraverseAllPairs(nums: number[]): boolean {
const n = nums.length;
if (n === 1) return true;
const p = Array.from({length: n}, (_, i) => i);
const find = (x: number): number => p[x] === x ? x : (p[x] = find(p[x]));
const primeToIdx = new Map<number, number>();
function factor(x: number): number[] {
const res: number[] = [];
for (let d = 2; d * d <= x; d++) {
if (x % d === 0) {
res.push(d);
while (x % d === 0) x /= d;
}
}
if (x > 1) res.push(x);
return res;
}
for (let i = 0; i < n; i++) {
for (const pf of factor(nums[i])) {
if (primeToIdx.has(pf)) {
p[find(i)] = find(primeToIdx.get(pf)!);
} else {
primeToIdx.set(pf, i);
}
}
}
const root = find(0);
for (let i = 1; i < n; i++) {
if (find(i) !== root) return false;
}
return true;
}
}
Complexity
- ⏰ Time complexity:
O(n * sqrt(m))— For each number, we factorize it up to sqrt(m), where m is the max value in nums. - 🧺 Space complexity:
O(n + m)— For DSU and prime-to-index mapping.