Is Partition Array Such That Maximum Difference Is K easy or hard?

The problem is rated Medium but conceptually straightforward once you recognize the greedy insight. Many candidates initially overcomplicate it with dynamic programming or brute force. Sorting followed by a linear greedy scan leads directly to the optimal solution.

Partition Array Such That Maximum Difference Is K Python/Java solution

The typical Python or Java implementation sorts the array and iterates once while tracking the start of the current group. If nums[i] - start > k, increment the group count and reset start to nums[i]. The logic is short, usually under 10–15 lines of code.

How to solve Partition Array Such That Maximum Difference Is K in O(n)?

Pure O(n) time is not achievable for arbitrary input because you must know the relative ordering of elements. The standard solution sorts the array first in O(n log n), then performs a linear greedy scan. If the input array were already sorted, the grouping step alone would run in O(n).

What is the best approach for Partition Array Such That Maximum Difference Is K?

The best approach is a greedy strategy after sorting the array. Sort the numbers, start a group with the smallest element, and keep adding elements until the difference between the current value and the group start exceeds k. Once it exceeds k, start a new group. This guarantees the minimum number of partitions with O(n log n) time complexity.

Is Partition Array Such That Maximum Difference Is K asked at Google/Amazon/Meta?

Array partitioning and greedy range grouping problems appear frequently in interviews at companies like Amazon, Google, and Meta. While this exact problem may vary, the pattern of sorting first and greedily forming valid groups is a common interview concept.

What data structure is used in Partition Array Such That Maximum Difference Is K?

The solution primarily uses arrays along with sorting and simple pointer variables. Some explanations frame the logic as a sliding window using two pointers on the sorted array. No advanced data structures like heaps or hash maps are required.

What is the time complexity of Partition Array Such That Maximum Difference Is K?

The optimal solution runs in O(n log n) time because the array must be sorted first. After sorting, a single linear scan groups elements using a greedy rule, which takes O(n). Space complexity is O(1) if the sort is done in-place.

Description Hints 3 Solution Video 18:25

2294. Partition Array Such That Maximum Difference Is K

Medium81.8% acceptance2 companies

You are given an integer array nums and an integer k. You may partition nums into one or more subsequences such that each element in nums appears in exactly one of the subsequences.

Return the minimum number of subsequences needed such that the difference between the maximum and minimum values in each subsequence is at most k.

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

Example 1:

Input: nums = [3,6,1,2,5], k = 2
Output: 2
Explanation:
We can partition nums into the two subsequences [3,1,2] and [6,5].
The difference between the maximum and minimum value in the first subsequence is 3 - 1 = 2.
The difference between the maximum and minimum value in the second subsequence is 6 - 5 = 1.
Since two subsequences were created, we return 2. It can be shown that 2 is the minimum number of subsequences needed.

Example 2:

Input: nums = [1,2,3], k = 1
Output: 2
Explanation:
We can partition nums into the two subsequences [1,2] and [3].
The difference between the maximum and minimum value in the first subsequence is 2 - 1 = 1.
The difference between the maximum and minimum value in the second subsequence is 3 - 3 = 0.
Since two subsequences were created, we return 2. Note that another optimal solution is to partition nums into the two subsequences [1] and [2,3].

Example 3:

Input: nums = [2,2,4,5], k = 0
Output: 3
Explanation:
We can partition nums into the three subsequences [2,2], [4], and [5].
The difference between the maximum and minimum value in the first subsequences is 2 - 2 = 0.
The difference between the maximum and minimum value in the second subsequences is 4 - 4 = 0.
The difference between the maximum and minimum value in the third subsequences is 5 - 5 = 0.
Since three subsequences were created, we return 3. It can be shown that 3 is the minimum number of subsequences needed.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁵
0 <= k <= 10⁵

Asked at 2 companies

Amazon

Google

Topics

Hints open one at a time

Hint 1

Which values in each subsequence matter? The only values that matter are the maximum and minimum values.

Hint 2

Let the maximum and minimum values of a subsequence be Max and Min. It is optimal to place all values in between Max and Min in the original array in the same subsequence as Max and Min.

Hint 3

Sort the array.

18:25

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