Description
You are given a positive integer array nums.
Partition nums into two arrays,Β nums1 and nums2, such that:
- Each element of the array
numsbelongs to either the arraynums1or the arraynums2. - 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.
Β
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.
Β
Constraints:
2 <= nums.length <= 1051 <= nums[i] <= 109
Solution
Python3
class Solution:
def findValueOfPartition(self, nums: List[int]) -> int:
N = len(nums)
nums.sort()
res = inf
for i in range(N - 1):
res = min(res, nums[i + 1] - nums[i])
return res