Leetcode #1998: GCD Sort of an Array
In this guide, we solve Leetcode #1998 GCD Sort of an Array in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Union Find, Array, Math, Number Theory, Sorting
Intuition
We need to merge components and check connectivity efficiently.
Union-Find supports near-constant-time merges and finds.
Approach
Initialize each node as its own parent and union pairs as you scan.
Use path compression to keep operations fast.
Steps:
- Initialize parent arrays.
- Union related nodes.
- Use find to check connectivity.
Example
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]
Python Solution
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
Complexity
The time complexity is Near O(n) (amortized). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.