README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2700-2799/2740.Find%20the%20Value%20of%20the%20Partition/README_EN.md	1301	Weekly Contest 350 Q2	Array Sorting

2740. Find the Value of the Partition

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Description

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

Solutions

Solution 1: Sorting

The problem requires us to minimize the partition value. Therefore, we can sort the array and then take the minimum difference between two adjacent numbers.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array.

Python3

class Solution:
    def findValueOfPartition(self, nums: List[int]) -> int:
        nums.sort()
        return min(b - a for a, b in pairwise(nums))

Java

class Solution {
    public int findValueOfPartition(int[] nums) {
        Arrays.sort(nums);
        int ans = 1 << 30;
        for (int i = 1; i < nums.length; ++i) {
            ans = Math.min(ans, nums[i] - nums[i - 1]);
        }
        return ans;
    }
}

C++

class Solution {
public:
    int findValueOfPartition(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int ans = 1 << 30;
        for (int i = 1; i < nums.size(); ++i) {
            ans = min(ans, nums[i] - nums[i - 1]);
        }
        return ans;
    }
};

Go

func findValueOfPartition(nums []int) int {
	sort.Ints(nums)
	ans := 1 << 30
	for i, x := range nums[1:] {
		ans = min(ans, x-nums[i])
	}
	return ans
}

TypeScript

function findValueOfPartition(nums: number[]): number {
    nums.sort((a, b) => a - b);
    let ans = Infinity;
    for (let i = 1; i < nums.length; ++i) {
        ans = Math.min(ans, Math.abs(nums[i] - nums[i - 1]));
    }
    return ans;
}

Rust

impl Solution {
    pub fn find_value_of_partition(mut nums: Vec<i32>) -> i32 {
        nums.sort();
        let mut ans = i32::MAX;
        for i in 1..nums.len() {
            ans = ans.min(nums[i] - nums[i - 1]);
        }
        ans
    }
}