Introduction

In this blog post, we will explore how to solve the LeetCode problem "2294" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the solution:

• Sort the input array nums.
• Iterate through the sorted array, greedily forming subsequences. Start a new subsequence whenever the difference between the current element and the minimum element of the current subsequence exceeds k.

• The number of subsequences formed is the answer.

• Runtime Complexity: O(n log n) due to sorting. Storage Complexity: O(1) excluding the space used for sorting (which might be O(log n) or O(n) depending on the sorting algorithm implementation).

Code

    def minSubsequences(nums, k):
    nums.sort()
    count = 0
    start = 0
    for i in range(len(nums)):
        if nums[i] - nums[start] > k:
            count += 1
            start = i
    return count + 1