3241. Divide Array Into Arrays With Max Difference¶

Difficulty: Medium

3241. Divide Array Into Arrays With Max Difference

Medium

You are given an integer array nums of size n where n is a multiple of 3 and a positive integer k.

Divide the array nums into n / 3 arrays of size 3 satisfying the following condition:

The difference between any two elements in one array is less than or equal to k.

Return a 2D array containing 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:

The difference between any two elements in each array is less than or equal to 2.

Example 2:

Input: nums = [2,4,2,2,5,2], k = 2

Output: []

Explanation:

Different ways to divide nums into 2 arrays of size 3 are:

[[2,2,2],[2,4,5]] (and its permutations)
[[2,2,4],[2,2,5]] (and its permutations)

Because there are four 2s there will be an array with the elements 2 and 5 no matter how we divide it. since 5 - 2 = 3 > k, the condition is not satisfied and so there is no valid division.

Example 3:

Input: nums = [4,2,9,8,2,12,7,12,10,5,8,5,5,7,9,2,5,11], k = 14

Output: [[2,2,12],[4,8,5],[5,9,7],[7,8,5],[5,9,10],[11,12,2]]

Explanation:

The difference between any two elements in each array is less than or equal to 14.

Constraints:

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

Solution¶

class Solution {
    public int[][] divideArray(int[] nums, int k) {
        int n = nums.length;
        int ans[][] = new int[n / 3][3];
        if (n % 3 != 0)
            return ans;
        Arrays.sort(nums);
        ArrayList<ArrayList<Integer>> res = new ArrayList<>();
        int i = 0;
        ArrayList<Integer> temp = new ArrayList<>();
        while(i < n) {
            if(temp.size() == 3) {
                res.add(new ArrayList<>(temp));
                temp.clear();
            }
            else 
                temp.add(nums[i++]);
        }
        if(temp.size() == 3) 
            res.add(new ArrayList<>(temp));        
        for(int l = 0; l < res.size(); l++) {
            for(int j = 0; j < res.get(l).size(); j++) 
                ans[l][j] = res.get(l).get(j);
        }
        boolean found = false;
        for(ArrayList<Integer> current : res) {
            int first = current.get(0), second = current.get(1), third = current.get(2);
            if(Math.abs(first - second) <= k && Math.abs(second - third) <= k && Math.abs(first - third) <= k) 
                continue;
            else 
                found = true;
            if (found == true)
                break;
        }
        if(found) 
            return new int[][]{};
        return ans;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here