#2966

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Divide Array Into Arrays With Max Difference

Divide an array into subarrays with a maximum element difference under a given threshold using a greedy approach.

Problem Statement

You are given an integer array nums where the length of the array, n, is a multiple of 3. The task is to divide this array into n / 3 subarrays, each containing exactly three elements. For each subarray, the absolute difference between the maximum and minimum element must be less than or equal to a given integer k.

Return a 2D array of these subarrays. If it is impossible to form such subarrays, return an empty array. If multiple valid subarray divisions exist, return any one of them.

Examples

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]]

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: []

Different ways to divide nums into 2 arrays of size 3 are: 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: [[2,2,2],[4,5,5],[5,5,7],[7,8,8],[9,9,10],[11,12,12]]

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 Approach

Sort the Array

Start by sorting the input array nums. Sorting helps easily group numbers with a minimal difference. The greedy choice of sorting ensures that elements closer together are placed together in the subarrays.

Greedy Subarray Formation

After sorting, try to form subarrays of 3 elements sequentially. For each potential subarray, check if the difference between the maximum and minimum element in the subarray is less than or equal to k. If any subarray doesn't meet this condition, return an empty array.

Return the Result

If all subarrays are valid, return the 2D array. Otherwise, if at any point the subarray conditions are violated, return an empty array. The solution must run in O(N log N) time due to the sorting step.

Complexity Analysis

Metric	Value
Time	O(N\cdot logN)
Space	O(N)

The time complexity is dominated by the sorting step, which is O(N log N). The space complexity is O(N) due to the storage of the result in a 2D array.

What Interviewers Usually Probe

The candidate understands greedy algorithms and can apply them to array manipulation.
The candidate recognizes the importance of sorting in solving the problem.
The candidate can efficiently handle edge cases, such as when no valid subarray division is possible.

Common Pitfalls or Variants

Common pitfalls

Failing to sort the array, leading to incorrect subarray divisions.
Not checking the subarray's minimum and maximum values correctly, resulting in invalid solutions.
Overlooking edge cases where no valid solution exists.

Follow-up variants

Change the subarray size from 3 to another fixed value, keeping the difference condition.
Adjust the threshold k to a larger or smaller value and observe how the solution adapts.
Consider other array types, such as arrays with duplicates or large ranges of numbers.

FAQ

What is the main approach to solving the "Divide Array Into Arrays With Max Difference" problem?

The main approach is to use a greedy strategy combined with sorting. After sorting the array, you try to form subarrays of 3 elements and check if the maximum and minimum element differences are within the allowed threshold.

Why is sorting necessary in the "Divide Array Into Arrays With Max Difference" problem?

Sorting ensures that elements with smaller differences are placed together, making it easier to form valid subarrays. Without sorting, finding valid subarrays would be much harder.

What happens if the subarray conditions are violated in this problem?

If at any point a subarray's maximum and minimum difference exceeds the threshold k, you should immediately return an empty array, as it's not possible to divide the array under the given conditions.

Can there be multiple valid answers for the "Divide Array Into Arrays With Max Difference" problem?

Yes, multiple valid subarray divisions can exist. The problem allows returning any valid solution.

What is the time complexity of the "Divide Array Into Arrays With Max Difference" problem?

The time complexity is O(N log N) due to the sorting step, where N is the length of the array.

terminal

Solution

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.

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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

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