How do you solve Partition Array Such That Maximum Difference Is K on LeetCode?

Partition Array Such That Maximum Difference Is K (LeetCode #2294) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: The maximum and minimum values in a subsequence are critical for determining if it can be formed.

What is the time complexity of Partition Array Such That Maximum Difference Is K?

The optimal solution for Partition Array Such That Maximum Difference Is K runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to the sorting step, while the linear pass through the array is O(n). This is efficient compared to the brute force approach.

What is the brute force solution for Partition Array Such That Maximum Difference Is K?

The brute force approach for Partition Array Such That Maximum Difference Is K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries to generate all possible subsequences and checks the maximum and minimum values for each. This is straightforward but inefficient, as it doesn't leverage any structure in the data. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Partition Array Such That Maximum Difference Is K?

Partition Array Such That Maximum Difference Is K uses the following concepts: Array, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting, Two Pointers.

What are the interview tips for Partition Array Such That Maximum Difference Is K?

Always consider sorting as a first step for problems involving ranges or differences. Think about how to minimize the number of operations — in this case, minimizing subsequences. Practice explaining your thought process clearly, as it helps in interviews.

What are common mistakes when solving Partition Array Such That Maximum Difference Is K?

Failing to sort the array before processing, which leads to incorrect subsequence grouping. Not properly checking the difference condition when iterating through the array.

#2294

Partition Array Such That Maximum Difference Is K

Medium

ArrayGreedySortingGreedySortingTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

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Intuition

Time O(n log n)Space O(1)

By sorting the array first, we can efficiently group elements together. The maximum difference condition can be checked in a linear pass, allowing us to minimize the number of subsequences needed.

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Algorithm

5 steps

1Step 1: Sort the array nums.
2Step 2: Initialize a counter for subsequences and set the first element as the start of the first subsequence.
3Step 3: Iterate through the sorted array, and for each element, check if it can fit into the current subsequence (i.e., if the difference with the start is <= k).
4Step 4: If it can't fit, increment the subsequence counter and start a new subsequence with the current element.
5Step 5: Return the total count of subsequences.

solution.py10 lines

1# Full working Python code
2def min_partitions(nums, k):
3    nums.sort()
4    count = 1
5    start = nums[0]
6    for num in nums:
7        if num - start > k:
8            count += 1
9            start = num
10    return count

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the linear pass through the array is O(n). This is efficient compared to the brute force approach.

1The maximum and minimum values in a subsequence are critical for determining if it can be formed.
2Sorting the array allows us to efficiently group elements that can fit within the difference constraint.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.