LeetCode #1998 — HARD

GCD Sort of an Array

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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.

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Union-Find

Example 1

[7,21,3]

Example 2

[5,2,6,2]

Example 3

[10,5,9,3,15]

Core Insight

What unlocks the optimal approach

Can we build a graph with all the prime numbers and the original array?
We can use union-find to determine which indices are connected (i.e., which indices can be swapped).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1998: GCD Sort of an Array
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) {
                f.computeIfAbsent(j, k -> new ArrayList<>()).add(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];
    }
}

// Accepted solution for LeetCode #1998: GCD Sort of an Array
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]
}

# Accepted solution for LeetCode #1998: GCD Sort of an Array
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

// Accepted solution for LeetCode #1998: GCD Sort of an Array
/**
 * [1998] GCD Sort of an Array
 *
 * 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 = [<u>21</u>,<u>7</u>,3]
 * - Swap 21 and 3 because gcd(21,3) = 3. nums = [<u>3</u>,7,<u>21</u>]
 *
 * 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 = [<u>15</u>,5,9,3,<u>10</u>]
 * - Swap 15 and 3 because gcd(15,3) = 3. nums = [<u>3</u>,5,9,<u>15</u>,10]
 * - Swap 10 and 15 because gcd(10,15) = 5. nums = [3,5,9,<u>10</u>,<u>15</u>]
 *
 *  
 * Constraints:
 *
 * 	1 <= nums.length <= 3 * 10^4
 * 	2 <= nums[i] <= 10^5
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/gcd-sort-of-an-array/
// discuss: https://leetcode.com/problems/gcd-sort-of-an-array/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn gcd_sort(nums: Vec<i32>) -> bool {
        false
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_1998_example_1() {
        let nums = vec![7, 21, 3];

        let result = true;

        assert_eq!(Solution::gcd_sort(nums), result);
    }

    #[test]
    #[ignore]
    fn test_1998_example_2() {
        let nums = vec![5, 2, 6, 2];

        let result = false;

        assert_eq!(Solution::gcd_sort(nums), result);
    }

    #[test]
    #[ignore]
    fn test_1998_example_3() {
        let nums = vec![10, 5, 9, 3, 15];

        let result = true;

        assert_eq!(Solution::gcd_sort(nums), result);
    }
}

// Accepted solution for LeetCode #1998: GCD Sort of an Array
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1998: GCD Sort of an Array
// 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) {
//                 f.computeIfAbsent(j, k -> new ArrayList<>()).add(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];
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(α(n))

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND

O(α(n)) time

O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.