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Leetcode #2740: Find the Value of the Partition

In this guide, we solve Leetcode #2740 Find the Value of the Partition 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 a positive integer array nums. Partition nums into two arrays, nums1 and nums2, such that: Each element of the array nums belongs to either the array nums1 or the array nums2.

Quick Facts

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

Intuition

Sorting reveals structure that is hard to see in the original order.

Once sorted, a linear scan is usually enough to compute the answer.

Approach

Sort the data and sweep through it while maintaining a small state.

This keeps the logic straightforward and reliable.

Steps:

  • Sort the data.
  • Scan in order while maintaining state.
  • Update the best answer.

Example

Input: nums = [1,3,2,4] Output: 1 Explanation: We can partition the array nums into nums1 = [1,2] and nums2 = [3,4]. - The maximum element of the array nums1 is equal to 2. - The minimum element of the array nums2 is equal to 3. The value of the partition is |2 - 3| = 1. It can be proven that 1 is the minimum value out of all partitions.

Python Solution

class Solution: def findValueOfPartition(self, nums: List[int]) -> int: nums.sort() return min(b - a for a, b in pairwise(nums))

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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