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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 <= 10`

^{5}`0 <= nums[i] <= 10`

^{5}`0 <= k <= 10`

^{5}

```
class Solution {
public int partitionArray(int[] nums, int k) {
Arrays.sort(nums);
int c = 1, prev = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] - nums[prev] <= k) continue;
c++; prev = i;
}
return c;
}
}
```

```
class Solution {
public:
int partitionArray(vector<int>& nums, int k) {
int n(size(nums)), res(0);
sort(begin(nums), end(nums));
for (int i=0, j=0; i<n;) {
while (j < n and nums[j] - nums[i] <= k) j++;
res++;
i = j;
}
return res;
}
};
```

```
class Solution:
def partitionArray(self, nums: List[int], k: int) -> int:
nums.sort()
ans = 1
start = nums[0]
for i in range(1, len(nums)):
diff = nums[i] - start
if diff > k:
ans += 1
start = nums[i]
return ans
```

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