LeetCode-in-Java

2740. 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;

public class Solution {
    public int findValueOfPartition(int[] nums) {
        Arrays.sort(nums);
        int minDifference = Integer.MAX_VALUE;
        for (int i = 1; i < nums.length; i++) {
            int difference = nums[i] - nums[i - 1];
            if (difference < minDifference) {
                minDifference = difference;
            }
        }
        return minDifference;
    }
}

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