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Formatted question description: https://leetcode.ca/all/2294.html

2294. Partition Array Such That Maximum Difference Is K

• Difficulty: Medium.
• Related Topics: Array, Greedy, Sorting.
• Similar Questions: Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit.

Problem

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

Solution (Java, C++, Python)

• Java
• Python
• C++
• Go
• TypeScript

• class Solution {
      public int partitionArray(int[] nums, int k) {
          Arrays.sort(nums);
          int partitions = 1;
          int j = 0;
          for (int i = 1; i < nums.length; i++) {
              if (nums[i] - nums[j] > k) {
                  partitions++;
                  j = i;
              }
          }
          return partitions;
      }
  }
  
  ############
  
  class Solution {
      public int partitionArray(int[] nums, int k) {
          Arrays.sort(nums);
          int ans = 1, a = nums[0];
          for (int b : nums) {
              if (b - a > k) {
                  a = b;
                  ++ans;
              }
          }
          return ans;
      }
  }
  

• class Solution:
      def partitionArray(self, nums: List[int], k: int) -> int:
          nums.sort()
          ans, a = 1, nums[0]
          for b in nums:
              if b - a > k:
                  a = b
                  ans += 1
          return ans
  
  ############
  
  # 2294. Partition Array Such That Maximum Difference Is K
  # https://leetcode.com/problems/partition-array-such-that-maximum-difference-is-k/
  
  class Solution:
      def partitionArray(self, nums: List[int], k: int) -> int:
          n = len(nums)
          nums.sort()
          res = 1
          mmin = mmax = nums[0]
          
          for i in range(1, n):
              if nums[i] > mmax:
                  mmax = nums[i]
              
              if nums[i] < mmin:
                  mmin = nums[i]
              
              if mmax - mmin > k:
                  res += 1
                  mmax = mmin = nums[i]
          
          return res
  
  

• class Solution {
  public:
      int partitionArray(vector<int>& nums, int k) {
          sort(nums.begin(), nums.end());
          int ans = 1, a = nums[0];
          for (int& b : nums) {
              if (b - a > k) {
                  a = b;
                  ++ans;
              }
          }
          return ans;
      }
  };
  

• func partitionArray(nums []int, k int) int {
  	sort.Ints(nums)
  	ans, a := 1, nums[0]
  	for _, b := range nums {
  		if b-a > k {
  			a = b
  			ans++
  		}
  	}
  	return ans
  }
  

• function partitionArray(nums: number[], k: number): number {
      nums.sort((a, b) => a - b);
      let ans = 1;
      let a = nums[0];
      for (const b of nums) {
          if (b - a > k) {
              a = b;
              ++ans;
          }
      }
      return ans;
  }
  
  

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).

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