README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2200-2299/2294.Partition%20Array%20Such%20That%20Maximum%20Difference%20Is%20K/README_EN.md	1416	Weekly Contest 296 Q2	Greedy Array Sorting

2294. Partition Array Such That Maximum Difference Is K

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Description

You are given an integer array nums and an integer k. You may partition nums into one or more subsequences such that each element in nums appears in exactly one of the subsequences.

Return the minimum number of subsequences needed such that the difference between the maximum and minimum values in each subsequence is at most k.

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

 

Example 1:

Input: nums = [3,6,1,2,5], k = 2
Output: 2
Explanation:
We can partition nums into the two subsequences [3,1,2] and [6,5].
The difference between the maximum and minimum value in the first subsequence is 3 - 1 = 2.
The difference between the maximum and minimum value in the second subsequence is 6 - 5 = 1.
Since two subsequences were created, we return 2. It can be shown that 2 is the minimum number of subsequences needed.

Example 2:

Input: nums = [1,2,3], k = 1
Output: 2
Explanation:
We can partition nums into the two subsequences [1,2] and [3].
The difference between the maximum and minimum value in the first subsequence is 2 - 1 = 1.
The difference between the maximum and minimum value in the second subsequence is 3 - 3 = 0.
Since two subsequences were created, we return 2. Note that another optimal solution is to partition nums into the two subsequences [1] and [2,3].

Example 3:

Input: nums = [2,2,4,5], k = 0
Output: 3
Explanation:
We can partition nums into the three subsequences [2,2], [4], and [5].
The difference between the maximum and minimum value in the first subsequences is 2 - 2 = 0.
The difference between the maximum and minimum value in the second subsequences is 4 - 4 = 0.
The difference between the maximum and minimum value in the third subsequences is 5 - 5 = 0.
Since three subsequences were created, we return 3. It can be shown that 3 is the minimum number of subsequences needed.

 

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 105
• 0 <= k <= 105

Solutions

Solution 1: Greedy + Sorting

The problem requires dividing into subsequences, not subarrays, so the elements in a subsequence can be non-continuous. We can sort the array $\textit{nums}$. Assuming the first element of the current subsequence is $a$, the difference between the maximum and minimum values in the subsequence will not exceed $k$. Therefore, we can iterate through the array $\textit{nums}$. If the difference between the current element $b$ and $a$ is greater than $k$, then update $a$ to $b$ and increase the number of subsequences by 1. After the iteration, we can obtain the minimum number of subsequences, noting that the initial number of subsequences is $1$.

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 $\textit{nums}$.

Python3

class Solution:
    def partitionArray(self, nums: List[int], k: int) -> int:
        nums.sort()
        ans, a = 1, nums[0]
        for b in nums:
            if b - a > k:
                a = b
                ans += 1
        return ans

Java

class Solution {
    public int partitionArray(int[] nums, int k) {
        Arrays.sort(nums);
        int ans = 1, a = nums[0];
        for (int b : nums) {
            if (b - a > k) {
                a = b;
                ++ans;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int partitionArray(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());
        int ans = 1, a = nums[0];
        for (int& b : nums) {
            if (b - a > k) {
                a = b;
                ++ans;
            }
        }
        return ans;
    }
};

Go

func partitionArray(nums []int, k int) int {
	sort.Ints(nums)
	ans, a := 1, nums[0]
	for _, b := range nums {
		if b-a > k {
			a = b
			ans++
		}
	}
	return ans
}

TypeScript

function partitionArray(nums: number[], k: number): number {
    nums.sort((a, b) => a - b);
    let ans = 1;
    let a = nums[0];
    for (const b of nums) {
        if (b - a > k) {
            a = b;
            ++ans;
        }
    }
    return ans;
}