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Leetcode #2294: Partition Array Such That Maximum Difference Is K

In this guide, we solve Leetcode #2294 Partition Array Such That Maximum Difference Is K in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

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.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Greedy, Array, Sorting

Intuition

A locally optimal choice leads to a globally optimal result for this structure.

That means we can commit to decisions as we scan without backtracking.

Approach

Sort or preprocess if needed, then repeatedly take the best available local choice.

Maintain the minimal state necessary to validate the greedy decision.

Steps:

  • Sort or preprocess as needed.
  • Iterate and pick the best local option.
  • Track the current solution.

Example

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.

Python Solution

class Solution: def partitionArray(self, nums: List[int], k: int) -> int: nums.sort() ans, a = 1, nums[0] for b in nums: if b - a > k: a = b ans += 1 return ans

Complexity

The time complexity is O(n×log⁡n)O(n \times \log n)O(n×logn), and the space complexity is O(log⁡n)O(\log n)O(logn). The space complexity is O(log⁡n)O(\log n)O(logn).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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