How do you solve GCD Sort of an Array on LeetCode?

GCD Sort of an Array (LeetCode #1998) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Elements that share a GCD greater than 1 can be swapped, forming connected components.

What is the brute force solution for GCD Sort of an Array?

The brute force approach for GCD Sort of an Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible pairs of elements to see if they can be swapped based on their GCD. If we can swap elements to eventually sort the array, we return true; otherwise, we return false. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve GCD Sort of an Array?

GCD Sort of an Array uses the following concepts: Array, Math, Union-Find, Sorting, Number Theory. The recommended patterns to study are: Union-Find, Sorting, Graph Theory.

What are the interview tips for GCD Sort of an Array?

Explain your thought process clearly when discussing the union-find approach. Be prepared to discuss the time complexity of your solution. Practice implementing union-find as it is a common data structure in many problems.

What are common mistakes when solving GCD Sort of an Array?

Failing to recognize the need for a union-find structure. Not checking if all elements in a connected component can be sorted correctly.

#1998

GCD Sort of an Array

Hard

ArrayMathUnion-FindSortingNumber TheoryUnion-FindSortingGraph Theory

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal approach uses the Union-Find (Disjoint Set Union) data structure to group elements that can be swapped based on their GCD. This allows us to efficiently determine if we can sort the array by checking if all elements in the same group can be rearranged.

⚙️

Algorithm

3 steps

1Step 1: Create a Union-Find structure to group indices based on GCD connections.
2Step 2: For each pair of elements, if gcd(nums[i], nums[j]) > 1, union their indices.
3Step 3: For each group of connected indices, check if the elements can be sorted within that group.

solution.py31 lines

1class UnionFind:
2    def __init__(self, n):
3        self.parent = list(range(n))
4    def find(self, x):
5        if self.parent[x] != x:
6            self.parent[x] = self.find(self.parent[x])
7        return self.parent[x]
8    def union(self, x, y):
9        rootX = self.find(x)
10        rootY = self.find(y)
11        if rootX != rootY:
12            self.parent[rootY] = rootX
13
14
15def canSort(nums):
16    n = len(nums)
17    uf = UnionFind(n)
18    for i in range(n):
19        for j in range(i + 1, n):
20            if math.gcd(nums[i], nums[j]) > 1:
21                uf.union(i, j)
22    groups = {}
23    for i in range(n):
24        root = uf.find(i)
25        if root not in groups:
26            groups[root] = []
27        groups[root].append(nums[i])
28    for group in groups.values():
29        if sorted(group) != sorted(group):
30            return False
31    return True

ℹ

Complexity note: The complexity is dominated by the union-find operations and sorting the groups, leading to a more efficient solution than the brute force approach.

1Elements that share a GCD greater than 1 can be swapped, forming connected components.
2Sorting within each connected component must yield the same result as sorting the entire array.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.