2740

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

Related Topics:
Array, Sorting

Solution 1.

// OJ: https://leetcode.com/problems/find-the-value-of-the-partition
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1) if we allow sorting the array in place.
class Solution {
public:
    int findValueOfPartition(vector<int>& A) {
        int N = A.size(), ans = INT_MAX;
        sort(begin(A), end(A));
        for (int i = 1; i < N; ++i) {
            ans = min(ans, A[i] - A[i - 1]);
        }
        return ans;
    }
};