Formatted question description: https://leetcode.ca/all/1998.html

1998. GCD Sort of an Array

Level

Hard

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

Explanation: 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 * 10^4
2 <= nums[i] <= 10^5

Solution

Use union find to partition the numbers from 1 to 100000 into groups such that for any group, if a is in the group, then there always exists a number b such that gcd(a, b) > 1. Then generate a sorted version of nums, which is sorted. If for any 0 <= i < nums.length, nums[i] and sorted[i] belong to the same group, then return true. Otherwise, return false.

class Solution {
    public boolean gcdSort(int[] nums) {
        int maxNum = 100000;
        int[] parent = new int[maxNum + 1];
        boolean[] exists = new boolean[maxNum + 1];
        int length = nums.length;
        for (int i = 0; i < length; i++)
            exists[nums[i]] = true;
        for (int i = 1; i <= maxNum; i++)
            parent[i] = i;
        for (int i = 1; i <= maxNum; i++) {
            if (exists[i])
                divide(parent, i);
        }
        int[] sorted = new int[length];
        System.arraycopy(nums, 0, sorted, 0, length);
        Arrays.sort(sorted);
        for (int i = 0; i < length; i++) {
            if (find(parent, sorted[i]) != find(parent, nums[i]))
                return false;
        }
        return true;
    }

    public void union(int[] parent, int index1, int index2) {
        parent[find(parent, index1)] = find(parent, index2);
    }

    public int find(int[] parent, int index) {
        if (parent[index] != index)
            parent[index] = find(parent, parent[index]);
        return parent[index];
    }

    public int gcd(int num1, int num2) {
        while (num2 != 0) {
            int temp = num1;
            num1 = num2;
            num2 = temp % num2;
        }
        return num1;
    }

    public void divide(int[] parent, int x) {
        int y = x;
        for (int i = 2; i * i <= x; i++) {
            if (x % i == 0) {
                while (x % i == 0) {
                    x /= i;
                    union(parent, y, i);
                }
            }
        }
        if (x != 1)
            union(parent, y, x);
    }
}

// OJ: https://leetcode.com/problems/gcd-sort-of-an-array/
// Time: O(N * sqrt(M) + NlogN)
// Space: O(N * sqrt(M))
class UnionFind {
    vector<int> id;
public:
    UnionFind(int N) : id(N) {
        iota(begin(id), end(id), 0);
    }
    int find(int a) {
        return id[a] == a ? a : (id[a] = find(id[a]));
    }
    void connect(int a, int b) {
        id[find(a)] = find(b);
    }
};
class Solution {
    vector<int> factors(int n) {
        vector<int> ans;
        for (int d = 2; d * d <= n; ++d) {
            if (n % d) continue;
            ans.push_back(d);
            while (n % d == 0) n /= d; 
        }
        if (n > 1) ans.push_back(n);
        return ans;
    }
public:
    bool gcdSort(vector<int>& A) {
        unordered_map<int, int> m; // prime factor -> representative index
        int N = A.size();
        UnionFind uf(N);
        for (int i = 0; i < N; ++i) {
            for (int f : factors(A[i])) {
                if (m.count(f)) uf.connect(m[f], i);
                else m[f] = i;
            }
        }
        unordered_map<int, vector<int>> groups;
        for (int i = 0; i < N; ++i) {
            groups[uf.find(i)].push_back(i);
        }
        vector<int> sorted(N);
        for (auto &[_, before] : groups) {
            auto after = before;
            sort(begin(after), end(after), [&](int a, int b) { return A[a] < A[b]; });
            for (int i = 0; i < before.size(); ++i) {
                sorted[before[i]] = A[after[i]];
            }
        }
        for (int i = 1; i < N; ++i) {
            if (sorted[i] < sorted[i - 1]) return false;
        }
        return true;
    }
};

```
print("Todo!")
```

1998 - GCD Sort of an Array

1998. GCD Sort of an Array

Level

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Solution

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