Package g2201_2300.s2294_partition_array_such_that_maximum_difference_is_k

Class Solution

java.lang.Object
g2201_2300.s2294_partition_array_such_that_maximum_difference_is_k.Solution
public class Solution extends Object
2294 - Partition Array Such That Maximum Difference Is K\. Medium 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

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • partitionArray

      public int partitionArray(int[] nums, int k)