# Solving Leetcode Interviews in Seconds with AI: Find Minimum Operations to Make All Elements Divisible by Three


	# 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](https://www.chatmagic.app), 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
	```python
	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)
	```