#2294

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Partition Array Such That Maximum Difference Is K

Find the minimum number of subsequences required such that the difference between the maximum and minimum value in each subsequence is at most k.

array greedy sorting

Problem Statement

You are given an integer array nums and an integer k. You need to partition nums into one or more subsequences such that the difference between the maximum and minimum values in each subsequence does not exceed k. A subsequence is derived from the array by deleting some or no elements while preserving the order of the remaining elements.

Return the minimum number of subsequences needed to meet the constraint. The challenge is to identify how many subsequences are required, which can be achieved by using a greedy approach to group elements based on their maximum and minimum values.

Examples

Example 1

Input: nums = [3,6,1,2,5], k = 2

Output: 2

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

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

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 Approach

Greedy Choice with Sorting

Sort the array in ascending order. The greedy strategy is to iterate through the sorted array and try to add each element to an existing subsequence as long as the maximum difference in that subsequence does not exceed k. If adding an element would violate this constraint, start a new subsequence.

Max-Min Subsequence Invariant

By maintaining the maximum and minimum values for each subsequence, we ensure that the difference between them is always within the allowed range. This invariant is validated as we build subsequences, guaranteeing minimal subsequences are formed.

Efficient Partitioning

This approach results in O(n log n) time complexity due to sorting, followed by a linear scan of the array to partition it. This makes the solution efficient for large input sizes, as it avoids brute-force checks for every possible subsequence combination.

Complexity Analysis

Metric	Value
Time	O(n \log n)
Space	O(S_n)

The time complexity is O(n log n) due to the sorting step, followed by a linear pass (O(n)) to partition the array. The space complexity is O(n) for storing the subsequences during the process, which can be improved with space optimization if needed.

What Interviewers Usually Probe

The candidate demonstrates an understanding of greedy algorithms and array partitioning.
The candidate focuses on maintaining an invariant for maximum and minimum values in each subsequence.
The candidate optimizes the solution for time and space complexity, considering sorting and linear traversal.

Common Pitfalls or Variants

Common pitfalls

Overlooking the importance of sorting the array before partitioning.
Incorrectly partitioning the array without ensuring that the max-min difference condition holds.
Failing to consider the efficiency of the approach when working with large arrays.

Follow-up variants

What if the array is already sorted? How would the solution change?
What if the constraint for the max-min difference is much smaller, or zero?
How would the solution be impacted if negative values were allowed in the array?

FAQ

What is the main strategy to solve Partition Array Such That Maximum Difference Is K?

The main strategy is using a greedy approach combined with sorting the array. By iterating through the sorted array and checking the maximum and minimum differences in subsequences, we can minimize the number of partitions needed.

What is the time complexity of the solution?

The time complexity is O(n log n) due to the sorting step, followed by a linear scan of the array to partition it.

What is the greedy choice in this problem?

The greedy choice is to attempt to add each element to an existing subsequence, as long as the difference between the maximum and minimum values in the subsequence does not exceed k.

How does sorting help in solving this problem?

Sorting the array allows us to handle elements in increasing order, ensuring that we can easily group elements together while maintaining the constraint on the difference between the maximum and minimum values.

Can this problem be solved with a dynamic programming approach?

While dynamic programming could be applied, a greedy approach combined with sorting provides a more efficient solution to the problem, especially considering the time and space complexity constraints.

terminal

Solution

Solution 1: Greedy + Sorting

The problem requires dividing into subsequences, not subarrays, so the elements in a subsequence can be non-continuous. We can sort the array $\textit{nums}$. Assuming the first element of the current subsequence is $a$, the difference between the maximum and minimum values in the subsequence will not exceed $k$. Therefore, we can iterate through the array $\textit{nums}$. If the difference between the current element $b$ and $a$ is greater than $k$, then update $a$ to $b$ and increase the number of subsequences by 1. After the iteration, we can obtain the minimum number of subsequences, noting that the initial number of subsequences is $1$.

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

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