Introduction

In this blog post, we will explore how to solve the LeetCode problem "3397" 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 are allowed to perform the following operation on each element of the array at most once: Add an integer in the range [-k, k] to the element. Return the maximum possible number of distinct elements in nums after performing the operations. Example 1: Input: nums = [1,2,2,3,3,4], k = 2 Output: 6 Explanation: nums changes to [-1, 0, 1, 2, 3, 4] after performing operations on the first four elements. Example 2: Input: nums = [4,4,4,4], k = 1 Output: 3 Explanation: By adding -1 to nums[0] and 1 to nums[1], nums changes to [3, 5, 4, 4]. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 109 0 <= k <= 109

Explanation

Here's a breakdown of the approach and the Python code:

• Greedy Approach: For each number in the sorted array, try to make it distinct by adding the minimum possible value within the [-k, k] range such that it doesn't collide with previously considered distinct numbers.
• Set to Track Distinct Numbers: Use a set to efficiently keep track of the distinct numbers we've already created.

• Iterate and Adjust: Iterate through the sorted array, and for each number, check if a distinct number can be formed within its adjustable range [num - k, num + k]. Prefer to form the smallest distinct number possible in that range to leave room for subsequent elements.

• Runtime Complexity: O(n log n) due to sorting.

• Storage Complexity: O(n) in the worst case, due to the set.

Code

    def max_distinct_elements(nums, k):
    nums.sort()
    distinct_nums = set()
    count = 0
    for num in nums:
        min_val = num - k
        max_val = num + k

        found = False
        for val in range(min_val, max_val + 1):
            if val not in distinct_nums:
                distinct_nums.add(val)
                count += 1
                found = True
                break
        if not found and num not in distinct_nums and -k <=0 <=k:
            if (num) not in distinct_nums:
                distinct_nums.add(num)
                count+=1

    return count