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# Partition Array Such That Maximum Difference Is K LeetCode Solution

## Problem – Partition Array Such That Maximum Difference Is K LeetCode Solution

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`.

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 .
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  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], , and .
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`

### Partition Array Such That Maximum Difference Is K LeetCode Solution in Java

``````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;
}
}
``````

### Partition Array Such That Maximum Difference Is K LeetCode Solution in 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;
}
};
``````

### Partition Array Such That Maximum Difference Is K LeetCode Solution in Python

``````class Solution:
def partitionArray(self, nums: List[int], k: int) -> int:
nums.sort()
ans = 1
start = nums

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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