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

Master LeetCode problem 2294 (Partition Array Such That Maximum Difference Is K), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2294. Partition Array Such That Maximum Difference Is K

Problem Explanation

Explanation

To solve this problem, we can use a greedy approach. We will sort the input array nums, and then iteratively create subsequences by adding elements to the current subsequence until the difference between the maximum and minimum values in that subsequence exceeds k. At that point, we start a new subsequence and continue the process. The minimum number of subsequences will be equal to the number of times we start a new subsequence.

The steps are as follows:

  1. Sort the input array nums.
  2. Initialize a variable res to keep track of the number of subsequences.
  3. Initialize a variable start to store the starting index of the current subsequence.
  4. Iterate through the sorted nums array:
    • If the difference between the current element and the element at index start exceeds k, increment res and set start to the current index.
  5. Return the value of res.

The time complexity of this solution is O(nlogn) due to the sorting step, where n is the number of elements in the input array. The space complexity is O(1) as we are using a constant amount of extra space.

Solution Code

import java.util.Arrays;

class Solution {
    public int partitionArray(int[] nums, int k) {
        Arrays.sort(nums);
        int res = 1;
        int start = 0;
        
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] - nums[start] > k) {
                res++;
                start = i;
            }
        }
        
        return res;
    }
}

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