Is Minimum Operations to Make All Grid Elements Equal easy or hard?

The problem is categorized as Hard because it requires combining greedy reasoning with a 2D difference array technique. Many candidates initially attempt direct simulation, which becomes too slow. Recognizing that range updates can be tracked with a difference matrix is the key optimization.

Minimum Operations to Make All Grid Elements Equal Python/Java solution

The implementation typically builds a 2D difference matrix and iterates through the grid while maintaining prefix effects. When a mismatch is detected, the algorithm records the necessary update in the difference structure. The same logic translates directly across Python, Java, C++, Go, and TypeScript with identical O(m*n) complexity.

How to solve Minimum Operations to Make All Grid Elements Equal in O(n)?

Treat the grid as a 2D structure and process it with a difference array. Maintain cumulative prefix effects while iterating through the grid. When a cell's effective value does not match the desired value, compute the adjustment and record the update boundaries in the difference array. This ensures each operation is applied in constant time and the grid is processed in a single pass.

What is the best approach for Minimum Operations to Make All Grid Elements Equal?

The most efficient approach uses a 2D difference array combined with a greedy traversal. While scanning the grid from top-left to bottom-right, track the cumulative effect of prior operations using prefix accumulation. When the current cell differs from the target value, apply the required adjustment and record it in the difference matrix. This avoids repeatedly updating large submatrices and runs in O(m*n) time.

Is Minimum Operations to Make All Grid Elements Equal asked at Google/Amazon/Meta?

Problems involving grid transformations, prefix sums, and difference arrays appear frequently in interviews at companies like Google, Amazon, and Meta. Variants often require minimizing operations on matrices using efficient range updates or greedy traversal strategies.

What data structure is used in Minimum Operations to Make All Grid Elements Equal?

The key data structure is a 2D difference array, which allows constant-time updates for rectangular ranges. Combined with prefix accumulation, it efficiently propagates the impact of operations across the grid while scanning once.

What is the time complexity of Minimum Operations to Make All Grid Elements Equal?

The optimal solution runs in O(m*n) time where m and n are the grid dimensions. Each cell is processed exactly once while the difference array tracks the effect of operations. Space complexity is O(m*n) for storing the auxiliary difference structure used to propagate range updates.

3888. Minimum Operations to Make All Grid Elements Equal - Practice Online

You are given a 2D integer array grid of size m × n, and an integer k.

In one operation, you can:

Select any k x k submatrix of grid, and
Increment all elements inside that submatrix by 1.

Return the minimum number of operations required to make all elements in the grid equal. If it is not possible, return -1.

A submatrix (x1, y1, x2, y2) is a matrix that forms by choosing all cells matrix[x][y] where x1 <= x <= x2 and y1 <= y <= y2.

Example 1:

Input: grid = [[3,3,5],[3,3,5]], k = 2

Output: 2

Explanation:

Choose the left 2 x 2 submatrix (covering the first two columns) and apply the operation twice.

After 1 operation: [[4, 4, 5], [4, 4, 5]]
After 2 operations: [[5, 5, 5], [5, 5, 5]]

All elements become equal to 5. Thus, the minimum number of operations is 2.

Example 2:

Input: grid = [[1,2],[2,3]], k = 1

Output: 4

Explanation:

Since k = 1, each operation increments a single cell grid[i][j] by 1. To make all elements equal, the final value must be 3.

Increase grid[0][0] = 1 to 3, requiring 2 operations.
Increase grid[0][1] = 2 to 3, requiring 1 operation.
Increase grid[1][0] = 2 to 3, requiring 1 operation.

Thus, the minimum number of operations is 2 + 1 + 1 + 0 = 4.

Constraints:

1 <= m == grid.length <= 1000
1 <= n == grid[i].length <= 1000
-10⁵ <= grid[i][j] <= 10⁵
1 <= k <= min(m, n)