Description
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.length1 <= n <= 105nis a multiple of 31 <= nums[i] <= 1051 <= k <= 105
Solution
Python3
class Solution:
def divideArray(self, nums: List[int], k: int) -> List[List[int]]:
N = len(nums)
if N % 3 != 0: return []
nums.sort()
res = []
curr = []
for x in nums:
if len(curr) == 0:
curr.append(x)
else:
if x - curr[0] > k:
return []
curr.append(x)
if len(curr) == 3:
res.append(curr)
curr = []
return res