2294. Partition Array Such That Maximum Difference Is K LeetCode Solution

In this guide, you will get 2294. Partition Array Such That Maximum Difference Is K LeetCode Solution with the best time and space complexity. The solution to Partition Array Such That Maximum Difference Is K problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Partition Array Such That Maximum Difference Is K solution in C++
Partition Array Such That Maximum Difference Is K solution in Java
Partition Array Such That Maximum Difference Is K solution in Python
Additional Resources

2294. Partition Array Such That Maximum Difference Is K LeetCode Solution image

Problem Statement of Partition Array Such That Maximum Difference Is K

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

Complexity Analysis

Time Complexity: O(\texttt{sort})
Space Complexity: O(\texttt{sort})

2294. Partition Array Such That Maximum Difference Is K LeetCode Solution in C++

class Solution {
 public:
  int partitionArray(vector<int>& nums, int k) {
    ranges::sort(nums);

    int ans = 1;
    int mn = nums[0];

    for (int i = 1; i < nums.size(); ++i)
      if (mn + k < nums[i]) {
        ++ans;
        mn = nums[i];
      }

    return ans;
  }
};
/* code provided by PROGIEZ */

2294. 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 ans = 1;
    int mn = nums[0];

    for (int i = 1; i < nums.length; ++i)
      if (mn + k < nums[i]) {
        ++ans;
        mn = nums[i];
      }

    return ans;
  }
}
// code provided by PROGIEZ

2294. 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
    mn = nums[0]

    for i in range(1, len(nums)):
      if mn + k < nums[i]:
        ans += 1
        mn = nums[i]

    return ans
 # code by PROGIEZ