#### QUESTION:

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
```

#### EXPLANATION:

排序之后使用贪心算法, 用pre标记最小值, 然后循环查看当前值和pre的差, 如果大于k了, 就切一段下来, 把pre标记成当前值即可. 最后别忘记最后一段.

#### SOLUTION:

```
class Solution {
func partitionArray(_ nums: [Int], _ k: Int) -> Int {
let nums = nums.sorted()
var result = 0
var pre = nums[0]
for index in nums.indices {
if nums[index] - pre > k {
result += 1
pre = nums[index]
}
}
return result + 1
}
}
```