Introduction

In this blog post, we will explore how to solve the LeetCode problem "2598" 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 a 0-indexed integer array nums and an integer value. In one operation, you can add or subtract value from any element of nums. For example, if nums = [1,2,3] and value = 2, you can choose to subtract value from nums[0] to make nums = [-1,2,3]. The MEX (minimum excluded) of an array is the smallest missing non-negative integer in it. For example, the MEX of [-1,2,3] is 0 while the MEX of [1,0,3] is 2. Return the maximum MEX of nums after applying the mentioned operation any number of times. Example 1: Input: nums = [1,-10,7,13,6,8], value = 5 Output: 4 Explanation: One can achieve this result by applying the following operations: - Add value to nums[1] twice to make nums = [1,0,7,13,6,8] - Subtract value from nums[2] once to make nums = [1,0,2,13,6,8] - Subtract value from nums[3] twice to make nums = [1,0,2,3,6,8] The MEX of nums is 4. It can be shown that 4 is the maximum MEX we can achieve. Example 2: Input: nums = [1,-10,7,13,6,8], value = 7 Output: 2 Explanation: One can achieve this result by applying the following operation: - subtract value from nums[2] once to make nums = [1,-10,0,13,6,8] The MEX of nums is 2. It can be shown that 2 is the maximum MEX we can achieve. Constraints: 1 <= nums.length, value <= 105 -109 <= nums[i] <= 109

Explanation

Here's a breakdown of the solution:

• Residue Classes: The core idea is to group numbers based on their remainders when divided by value. This is because adding or subtracting value from a number doesn't change its residue modulo value.
• Frequency Counting: For each residue class, we count how many numbers fall into that class. We are trying to form the sequence 0, 1, 2, 3... so each residue class must contain at least one number before we can proceed to the next number.

• Maximum MEX: The maximum MEX is determined by how many complete sequences (0, 1, 2...) we can form from the residue classes. The residue class with the fewest elements limits the length of the achievable MEX.

• Time Complexity: O(n), where n is the length of the input array nums.

• Space Complexity: O(value), where value is the input integer value.

Code

    def find_mex(nums, value):
    """
    Calculates the maximum MEX (minimum excluded) of nums after applying the mentioned operation any number of times.

    Args:
        nums: A 0-indexed integer array.
        value: An integer value.

    Returns:
        The maximum MEX of nums after applying the operation.
    """

    remainders = {}
    for num in nums:
        remainder = num % value
        if remainder < 0:
            remainder += value  # Ensure non-negative remainder
        remainders[remainder] = remainders.get(remainder, 0) + 1

    min_count = float('inf')
    for remainder in range(value):
        count = remainders.get(remainder, 0)
        min_count = min(min_count, count)

    return min_count * value + len(remainders) if min_count > 0 else len(remainders) if min_count == 0 else 0