2966. Divide Array Into Arrays With Max Difference

Description

You are given an integer array nums of size n and a positive integer k.

Divide the array into one or more arrays of size 3 satisfying the following conditions:

  • Each element of nums should be in exactly one array.
  • The difference between any two elements in one array is less than or equal to k.

Return a 2D array containing all the arrays. If it is impossible to satisfy the conditions, return an empty array. And if there are multiple answers, return any of them.

 

Example 1:

Input: nums = [1,3,4,8,7,9,3,5,1], k = 2
Output: [[1,1,3],[3,4,5],[7,8,9]]
Explanation: We can divide the array into the following arrays: [1,1,3], [3,4,5] and [7,8,9].
The difference between any two elements in each array is less than or equal to 2.
Note that the order of elements is not important.

Example 2:

Input: nums = [1,3,3,2,7,3], k = 3
Output: []
Explanation: It is not possible to divide the array satisfying all the conditions.

 

Constraints:

  • n == nums.length
  • 1 <= n <= 105
  • n is a multiple of 3.
  • 1 <= nums[i] <= 105
  • 1 <= k <= 105

Solutions

Solution 1: Sorting

First, we sort the array. Then, we take out three elements each time. If the difference between the maximum and minimum values of these three elements is greater than $k$, then the condition cannot be satisfied, and we return an empty array. Otherwise, we add the array composed of these three elements to the answer array.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.

Python Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

class Solution:
    def divideArray(self, nums: List[int], k: int) -> List[List[int]]:
        nums.sort()
        ans = []
        n = len(nums)
        for i in range(0, n, 3):
            t = nums[i : i + 3]
            if t[2] - t[0] > k:
                return []
            ans.append(t)
        return ans

Java Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15

class Solution {
    public int[][] divideArray(int[] nums, int k) {
        Arrays.sort(nums);
        int n = nums.length;
        int[][] ans = new int[n / 3][];
        for (int i = 0; i < n; i += 3) {
            int[] t = Arrays.copyOfRange(nums, i, i + 3);
            if (t[2] - t[0] > k) {
                return new int[][] {};
            }
            ans[i / 3] = t;
        }
        return ans;
    }
}

C++ Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16

class Solution {
public:
    vector<vector<int>> divideArray(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> ans;
        int n = nums.size();
        for (int i = 0; i < n; i += 3) {
            vector<int> t = {nums[i], nums[i + 1], nums[i + 2]};
            if (t[2] - t[0] > k) {
                return {};
            }
            ans.emplace_back(t);
        }
        return ans;
    }
};

Go Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12

func divideArray(nums []int, k int) [][]int {
	sort.Ints(nums)
	ans := [][]int{}
	for i := 0; i < len(nums); i += 3 {
		t := slices.Clone(nums[i : i+3])
		if t[2]-t[0] > k {
			return [][]int{}
		}
		ans = append(ans, t)
	}
	return ans
}

TypeScript Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12

function divideArray(nums: number[], k: number): number[][] {
    nums.sort((a, b) => a - b);
    const ans: number[][] = [];
    for (let i = 0; i < nums.length; i += 3) {
        const t = nums.slice(i, i + 3);
        if (t[2] - t[0] > k) {
            return [];
        }
        ans.push(t);
    }
    return ans;
}