# 1998. GCD Sort of an Array

## Description

You are given an integer array nums, and you can perform the following operation any number of times on nums:

• Swap the positions of two elements nums[i] and nums[j] if gcd(nums[i], nums[j]) > 1 where gcd(nums[i], nums[j]) is the greatest common divisor of nums[i] and nums[j].

Return true if it is possible to sort nums in non-decreasing order using the above swap method, or false otherwise.

Example 1:

Input: nums = [7,21,3]
Output: true
Explanation: We can sort [7,21,3] by performing the following operations:
- Swap 7 and 21 because gcd(7,21) = 7. nums = [21,7,3]
- Swap 21 and 3 because gcd(21,3) = 3. nums = [3,7,21]


Example 2:

Input: nums = [5,2,6,2]
Output: false
Explanation: It is impossible to sort the array because 5 cannot be swapped with any other element.


Example 3:

Input: nums = [10,5,9,3,15]
Output: true
We can sort [10,5,9,3,15] by performing the following operations:
- Swap 10 and 15 because gcd(10,15) = 5. nums = [15,5,9,3,10]
- Swap 15 and 3 because gcd(15,3) = 3. nums = [3,5,9,15,10]
- Swap 10 and 15 because gcd(10,15) = 5. nums = [3,5,9,10,15]


Constraints:

• 1 <= nums.length <= 3 * 104
• 2 <= nums[i] <= 105

## Solutions

Union find.

• class Solution {
private int[] p;

public boolean gcdSort(int[] nums) {
int n = 100010;
p = new int[n];
Map<Integer, List<Integer>> f = new HashMap<>();
for (int i = 0; i < n; ++i) {
p[i] = i;
}
int mx = 0;
for (int num : nums) {
mx = Math.max(mx, num);
}
for (int i = 2; i <= mx; ++i) {
if (f.containsKey(i)) {
continue;
}
for (int j = i; j <= mx; j += i) {
}
}
for (int i : nums) {
for (int j : f.get(i)) {
p[find(i)] = find(j);
}
}
int[] s = new int[nums.length];
System.arraycopy(nums, 0, s, 0, nums.length);
Arrays.sort(s);
for (int i = 0; i < nums.length; ++i) {
if (s[i] != nums[i] && find(nums[i]) != find(s[i])) {
return false;
}
}
return true;
}

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

• class Solution {
public:
vector<int> p;

bool gcdSort(vector<int>& nums) {
int n = 100010;
p.resize(n);
for (int i = 0; i < n; ++i) p[i] = i;
int mx = 0;
for (auto num : nums) mx = max(mx, num);
unordered_map<int, vector<int>> f;
for (int i = 2; i <= mx; ++i) {
if (!f[i].empty()) continue;
for (int j = i; j <= mx; j += i) f[j].push_back(i);
}
for (int i : nums) {
for (int j : f[i]) p[find(i)] = find(j);
}
vector<int> s = nums;
sort(s.begin(), s.end());
for (int i = 0; i < nums.size(); ++i) {
if (s[i] != nums[i] && find(s[i]) != find(nums[i])) return false;
}
return true;
}

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

• class Solution:
def gcdSort(self, nums: List[int]) -> bool:
n = 10**5 + 10
p = list(range(n))
f = defaultdict(list)
mx = max(nums)
for i in range(2, mx + 1):
if f[i]:
continue
for j in range(i, mx + 1, i):
f[j].append(i)

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

for i in nums:
for j in f[i]:
p[find(i)] = find(j)

s = sorted(nums)
for i, num in enumerate(nums):
if s[i] != num and find(num) != find(s[i]):
return False
return True


• var p []int

func gcdSort(nums []int) bool {
n := 100010
p = make([]int, n)
for i := 0; i < n; i++ {
p[i] = i
}
mx := 0
for _, num := range nums {
mx = max(mx, num)
}
f := make([][]int, mx+1)
for i := 2; i <= mx; i++ {
if len(f[i]) > 0 {
continue
}
for j := i; j <= mx; j += i {
f[j] = append(f[j], i)
}
}
for _, i := range nums {
for _, j := range f[i] {
p[find(i)] = find(j)
}
}
s := make([]int, len(nums))
for i, num := range nums {
s[i] = num
}
sort.Ints(s)
for i, num := range nums {
if s[i] != num && find(s[i]) != find(num) {
return false
}
}
return true
}

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