Problem Description

You have a 2D grid of integers with dimensions m x n and a value x. Your goal is to transform this grid into a uni-value grid (where all elements are equal) using the minimum number of operations.

In each operation, you can:

• Add x to any element in the grid, OR
• Subtract x from any element in the grid

The problem asks you to find the minimum number of operations needed to make all elements in the grid equal. If it's impossible to achieve this, return -1.

Key insight: For it to be possible to make all elements equal through adding/subtracting x, all elements must have the same remainder when divided by x. This is because adding or subtracting x doesn't change an element's remainder modulo x.

For example, if x = 3:

• Starting with element 5 (remainder = 2), you can reach: 2, 5, 8, 11, ... (all have remainder 2)
• Starting with element 7 (remainder = 1), you can reach: 1, 4, 7, 10, ... (all have remainder 1)
• These two sets never overlap, so elements 5 and 7 can never be made equal using operations of adding/subtracting 3

The solution approach involves:

1. Checking if all elements have the same remainder when divided by x (if not, return -1)
2. Finding the optimal target value (the median of all elements minimizes total operations)
3. Calculating the total number of operations needed to transform each element to this target value

Intuition

Let's think about what happens when we add or subtract x from an element. If we have an element with value a, we can transform it to a + x, a - x, a + 2x, a - 2x, and so on. Notice that all reachable values from a can be written as a + k*x where k is any integer.

This means two elements a and b can be made equal only if there exist integers k1 and k2 such that: a + k1*x = b + k2*x

Rearranging: a - b = (k2 - k1)*x

This tells us that (a - b) must be divisible by x. In other words, a and b must have the same remainder when divided by x.

Since all elements need to become equal, they all must have the same remainder modulo x. If any element has a different remainder, it's impossible to make the grid uni-value.

Now, assuming all elements can be made equal, what value should we target? We need to minimize the total number of operations. Each element needs |element - target| / x operations to reach the target value.

The total operations would be: Σ |element - target| / x

Since x is constant, we're essentially minimizing Σ |element - target|, which is the sum of absolute deviations from the target.

From statistics, we know that the median minimizes the sum of absolute deviations. This is a key mathematical property: among all possible target values, the median requires the minimum total "distance" when distance is measured as absolute difference.

Therefore, our strategy is:

1. Verify all elements have the same remainder modulo x
2. Sort all elements and pick the median as our target
3. Calculate the total operations needed to transform each element to the median

Learn more about Math and Sorting patterns.

Solution Approach

The implementation follows a straightforward greedy approach based on our intuition:

Step 1: Flatten and Validate First, we need to check if transformation is possible. We iterate through the 2D grid and:

• Store the remainder of the first element: mod = grid[0][0] % x
• For each element v in the grid:
  • Check if v % x == mod (same remainder as the first element)
  • If any element has a different remainder, return -1 immediately
  • Otherwise, add the element to a 1D list nums for easier processing

Step 2: Find the Median After flattening the grid into a 1D array:

• Sort the array: nums.sort()
• Find the median element: mid = nums[len(nums) >> 1]
  • Note: len(nums) >> 1 is equivalent to len(nums) // 2 (right shift by 1 bit)
  • For even-length arrays, this gives us the lower middle element, which works as our target

Step 3: Calculate Total Operations With the median as our target value:

• For each element v in nums:
  • Calculate the absolute difference: |v - mid|
  • Divide by x to get the number of operations: |v - mid| // x
• Sum all these operations: sum(abs(v - mid) // x for v in nums)

The time complexity is O(m*n*log(m*n)) due to sorting, where m and n are the grid dimensions. The space complexity is O(m*n) for storing the flattened array.

This greedy approach guarantees the minimum number of operations because:

1. The median minimizes the sum of absolute deviations
2. Each element can only be changed by multiples of x, and we've verified all elements can reach the same value

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example Input:

• Grid: [[2, 8], [5, 11]]
• x = 3

Step 1: Flatten and Validate

First, we check if all elements have the same remainder when divided by x = 3:

• Element 2: 2 % 3 = 2
• Element 8: 8 % 3 = 2
• Element 5: 5 % 3 = 2
• Element 11: 11 % 3 = 2

All elements have remainder 2, so transformation is possible! We flatten them into: nums = [2, 8, 5, 11]

Step 2: Find the Median

Sort the array: nums = [2, 5, 8, 11]

Find the median:

• Array length is 4 (even)
• Median index: 4 // 2 = 2
• Median value: nums[2] = 8

Our target value is 8.

Step 3: Calculate Total Operations

For each element, calculate operations needed to reach target 8:

• Element 2 → 8:

  • Difference: |2 - 8| = 6
  • Operations: 6 / 3 = 2 (add 3 twice: 2 → 5 → 8)

• Element 5 → 8:

  • Difference: |5 - 8| = 3
  • Operations: 3 / 3 = 1 (add 3 once: 5 → 8)

• Element 8 → 8:

  • Difference: |8 - 8| = 0
  • Operations: 0 / 3 = 0 (already at target)

• Element 11 → 8:

  • Difference: |11 - 8| = 3
  • Operations: 3 / 3 = 1 (subtract 3 once: 11 → 8)

Total operations: 2 + 1 + 0 + 1 = 4

The minimum number of operations needed to transform the grid into a uni-value grid is 4.

Solution Implementation

Time and Space Complexity

Time Complexity: O((m × n) × log(m × n))

The algorithm performs the following operations:

• Iterating through the grid to flatten it into a list and check modulo conditions: O(m × n)
• Sorting the flattened list of m × n elements: O((m × n) × log(m × n))
• Computing the sum of absolute differences: O(m × n)

The dominant operation is the sorting step, resulting in an overall time complexity of O((m × n) × log(m × n)).

Space Complexity: O(m × n)

The algorithm uses additional space for:

• The nums list that stores all m × n elements from the grid: O(m × n)
• The sorting operation (Python's Timsort) requires O(m × n) space in the worst case

Therefore, the total space complexity is O(m × n), where m and n are the number of rows and columns of the grid, respectively.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Median Selection for Even-Length Arrays

A common misconception is that using the lower middle element for even-length arrays might not be optimal. Some might think they need to check both middle elements or calculate the average.

Why it's a pitfall: Developers might overcomplicate by trying both middle values or computing the average, which doesn't work since the target must be reachable through operations of adding/subtracting x.

Solution: Either middle element (lower or upper) gives the same minimum total operations for even-length arrays. The code correctly uses len(flattened_values) // 2 which automatically selects the lower middle element.

2. Forgetting to Check Remainder Consistency Early

Some implementations might first process all data (flattening, sorting) before checking if transformation is possible, leading to unnecessary computation when the answer is -1.

Why it's a pitfall: This wastes time and space processing data that will ultimately be discarded.

Solution: Check remainder consistency during the initial grid traversal (as shown in the code). Return -1 immediately upon finding a mismatch:

if value % x != reference_remainder:
    return -1  # Early termination

3. Integer Division vs Float Division

Using regular division (/) instead of integer division (//) when calculating operations can cause errors.

Why it's a pitfall:

• abs(value - median_value) / x returns a float
• This can lead to floating-point precision issues or type errors

Solution: Always use integer division (//) since the number of operations must be an integer:

total_operations = sum(abs(value - median_value) // x for value in flattened_values)

4. Handling Edge Case of x = 0

If x = 0, the code would crash with a division by zero error when checking remainders or calculating operations.

Why it's a pitfall: The problem statement might not explicitly exclude x = 0, but mathematically, you cannot add or subtract 0 to change values.

Solution: Add a guard clause at the beginning:

if x == 0:
    # Check if all elements are already equal
    first_val = grid[0][0]
    for row in grid:
        for value in row:
            if value != first_val:
                return -1
    return 0  # Already a uni-value grid

5. Not Understanding Why Median Minimizes Operations

Some might try using mean/average or mode instead of median as the target value.

Why it's a pitfall:

• Mean minimizes squared differences, not absolute differences
• Mode might not exist or might not be reachable through operations
• Only median minimizes the sum of absolute deviations

Solution: Stick with the median. It's mathematically proven that the median minimizes the sum of absolute distances to all points in a dataset.