Introduction

In this blog post, we will explore how to solve the LeetCode problem "3190" 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. In one operation, you can add or subtract 1 from any element of nums. Return the minimum number of operations to make all elements of nums divisible by 3. Example 1: Input: nums = [1,2,3,4] Output: 3 Explanation: All array elements can be made divisible by 3 using 3 operations: Subtract 1 from 1. Add 1 to 2. Subtract 1 from 4. Example 2: Input: nums = [3,6,9] Output: 0 Constraints: 1 <= nums.length <= 50 1 <= nums[i] <= 50

Explanation

Here's the breakdown of the solution:

• Calculate Remainders: Compute the remainder of each number in nums when divided by 3.
• Count Remainders: Count the occurrences of remainders 1 and 2. Remainder 0 requires no operations.

• Minimize Operations: If either remainder 1 or 2 has a count of 0, the minimum operations is the count of the other remainder. If both exist, pick the case where shifting all to remainder 0 needs the least operations.

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

• Storage Complexity: O(1).

Code

    def minOperations(nums):
    """
    Calculates the minimum number of operations to make all elements of nums divisible by 3.

    Args:
        nums: An integer array.

    Returns:
        The minimum number of operations.
    """

    count1 = 0
    count2 = 0

    for num in nums:
        remainder = num % 3
        if remainder == 1:
            count1 += 1
        elif remainder == 2:
            count2 += 1

    if count1 == 0:
        return count2
    if count2 == 0:
        return count1

    return min(count1, count2)