2740. Find the Value of the Partition

Problem

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 (Java)

class Solution {
  public int findValueOfPartition(int[] nums) {
    int n = nums.length, ans = Integer.MAX_VALUE;
    Arrays.sort(nums);

    for (var i=0; i < n-1; i++)
      ans = Math.min(ans, nums[i+1] - nums[i]);

    return ans;
  }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).