2845. Find The Value Of The Partition¶

Difficulty: Medium

2845. Find the Value of the Partition

Medium

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.
Both arrays are non-empty.
The value of the partition is minimized.

The value of the partition is |max(nums1) - min(nums2)|.

Here, max(nums1) denotes the maximum element of the array nums1, and min(nums2) denotes the minimum element of the array nums2.

Return the integer denoting the value of such partition.

Example 1:

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.

Example 2:

Input: nums = [100,1,10]
Output: 9
Explanation: We can partition the array nums into nums1 = [10] and nums2 = [100,1].
- The maximum element of the array nums1 is equal to 10.
- The minimum element of the array nums2 is equal to 1.
The value of the partition is |10 - 1| = 9.
It can be proven that 9 is the minimum value out of all partitions.

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Solution¶

import java.util.Arrays;

class Solution {
    public int findValueOfPartition(int[] nums) {
        int n = nums.length;
        int mini = Integer.MAX_VALUE;
        Arrays.sort(nums);
        for (int i = 0; i < n - 1; i++)
            mini = Math.min(mini, Math.max(nums[i], nums[i + 1]) - Math.min(nums[i], nums[i + 1]));
        return mini;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here