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2387 Partition Array Such That Maximum Difference Is K

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By Anurag
3 min read
2387 Partition Array Such That Maximum Difference Is K

Partition Array Such That Maximum Difference Is K image

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:

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**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:

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**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:

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**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:

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

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class Solution:
    def partitionArray(self, nums: List[int], k: int) -> int:
        """
        what if we went greedy, for every subsequence keep taking all
        elems , till max and min is satified
        and leave others
        is this union find ?
        find all nodes that at at most k 
        this approach is n2
        greedy does not work in case
        3,5,6,1,2-
        3,5 | 6 | 1,2
        5,6 | 3,1,2
        starting at index i, take longest subsequence, this greedy works
        /////// missed that sorting already works, I assumed soring will
        ////// only work for grps, rest was fine
        """
        nums.sort()
        grps = 1
        curr_min, curr_max = nums[0], nums[0]
        for k1 in nums:
            if k1 - curr_min <= k and k1 - curr_max <= k:
                curr_max = k1
                continue
            else:
                grps += 1
                curr_min = k1
                curr_max = k1
        return grps
leetcode
python
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