# Solving Leetcode Interviews in Seconds with AI: Minimum Operations to Make a Uni-Value Grid


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2033" 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 a 2D integer grid of size m x n and an integer x. In one operation, you can add x to or subtract x from any element in the grid. A uni-value grid is a grid where all the elements of it are equal. Return the minimum number of operations to make the grid uni-value. If it is not possible, return -1.   Example 1:   Input: grid = [[2,4],[6,8]], x = 2 Output: 4 Explanation: We can make every element equal to 4 by doing the following:  - Add x to 2 once. - Subtract x from 6 once. - Subtract x from 8 twice. A total of 4 operations were used.  Example 2:   Input: grid = [[1,5],[2,3]], x = 1 Output: 5 Explanation: We can make every element equal to 3.  Example 3:   Input: grid = [[1,2],[3,4]], x = 2 Output: -1 Explanation: It is impossible to make every element equal.    Constraints:  m == grid.length n == grid[i].length 1 <= m, n <= 105 1 <= m * n <= 105 1 <= x, grid[i][j] <= 104  

	# Explanation
	Here's the solution approach:

*   **Check Feasibility:** Determine if it's possible to make the grid uni-value. This is achievable only if the difference between any two elements is a multiple of `x`.
*   **Find Median:** Flatten the grid into a 1D array and find the median. The median is the optimal value to make all elements equal to in order to minimize the number of operations.
*   **Calculate Operations:** Calculate the number of operations needed to convert each element to the median value. Divide the absolute difference between each element and the median by `x` and sum the results.

*   **Runtime Complexity:** O(m\*n\*log(m\*n)), primarily due to sorting.
    **Storage Complexity:** O(m\*n) for storing the flattened grid.

	
	# Code
	```python
	def minOperations(grid, x):
    """
    Calculates the minimum operations to make the grid uni-value.

    Args:
        grid: A 2D integer grid.
        x: An integer.

    Returns:
        The minimum number of operations, or -1 if it's not possible.
    """

    nums = []
    for row in grid:
        nums.extend(row)

    n = len(nums)

    # Check if it's possible to make the grid uni-value
    for i in range(1, n):
        if abs(nums[i] - nums[0]) % x != 0:
            return -1

    # Find the median
    nums.sort()
    median = nums[n // 2]

    # Calculate the number of operations
    operations = 0
    for num in nums:
        operations += abs(num - median) // x

    return operations
	```