Problem Description

You are given a 2D matrix grid of size m x n. In one operation, you can change the value of any cell to any non-negative number. You need to perform some operations such that each cell grid[i][j] is:

• Equal to the cell below it, i.e. grid[i][j] == grid[i + 1][j] (if it exists).
• Different from the cell to its right, i.e. grid[i][j] != grid[i][j + 1] (if it exists).

Return the minimum number of operations needed.

Intuition

The goal is to transform the grid such that each column contains the same number and each row has distinct consecutive columns. Given that there are only 10 potential values (0 to 9) for each cell, we can use dynamic programming to efficiently find the solution.

1. Identify State: Define f[i][j] as the minimum number of operations required to make the first [0,..,i] columns valid, with the i-th column being filled with the number j.

2. Base Case: For the first column (i=0), we compute f[0][j] as m - cnt[j], where cnt[j] is the count of already existing cells in the column i=0 with number j.

3. State Transition: For each subsequent column i, calculate f[i][j] as the minimum over all previous column configurations that have a different number, plus the cost to make the entire column i equal to j. This can be expressed as:

   f[i][j] = min(f[i-1][k] + m - cnt[j]) for all k ≠ j

   Here, m - cnt[j] represents the number of changes needed for column i to make all elements j.

4. Result: After filling out the dynamic programming table, the solution is the minimum value of f[n-1][j], representing the minimum operations needed for the entire grid.

By exploring all possible column configurations while ensuring distinct column transformations, this method effectively minimizes the required operations.

Learn more about Dynamic Programming patterns.

Solution Approach

To solve this problem optimally, we employ a dynamic programming approach, which is well-suited for minimizing operations while considering constraints across rows and columns:

1. Initialize the DP Table: Create a 2D list f of size [n][10], where n is the number of columns, and each entry is initialized to infinity (inf). This table will store the minimum operations required to configure columns up to index i such that the i-th column is uniform with the value j.

   f = [[inf] * 10 for _ in range(n)]

2. Fill the First Column: For the first column, calculate and assign the number of changes necessary for each possible number j (from 0 to 9) to become uniform in that column.

   for j in range(10):
       f[0][j] = m - cnt[j]

   Here, cnt[j] counts how many times the number j already appears in the current column. m - cnt[j] gives the operations needed to make the whole column j.

3. Iterate Through Columns: For each subsequent column i, calculate the minimum operations required to make the column uniform with each number j considering different numbers in the previous column.

   for j in range(10):
       for k in range(10):
           if k != j:
               f[i][j] = min(f[i][j], f[i - 1][k] + m - cnt[j])

   The idea is to transition states only when adjacent columns have distinct values (k ≠ j). This guarantees the columns adhere to the prescribed conditions.

4. Compute the Final Result: Once the DP table is filled, the problem's answer is the minimum value found in the last column’s entries of table f.

   return min(f[n-1])

This solution efficiently leverages dynamic programming to track the least number of operation changes necessary while also ensuring the distinctness condition between consecutive columns is maintained.

Example Walkthrough

Let's consider a small example to illustrate the solution approach. Suppose we have the following grid of size 3 x 3:

grid = [
  [2, 3, 1],
  [2, 3, 1],
  [2, 3, 1]
]

Our task is to transform this grid so that:

• Each column contains the same number.
• Each row has distinct consecutive columns.

Step-by-Step Solution:

1. Initialize the DP Table:

   We create a DP table f of size [3][10] (since the grid has 3 columns, and each cell value can be between 0 and 9). Initialize each element to infinity (inf).

   f = [[inf] * 10 for _ in range(3)]

2. Fill the First Column:

   For the first column, we calculate the minimum operations needed to transform it into a uniform column for each number j from 0 to 9.

   Calculate cnt[j], the count of j in the first column, which in this case gives cnt[2] = 3 and others are 0. Thus:

   for j in range(10):
       f[0][j] = 3 - cnt[j]

   Here, f[0][2] = 0 because column 0 is already uniform with value 2. All others would require 3 changes.

3. Iterate Through Columns:

   Move to the second column, calculating minimum operations for each value j with distinct values from the first column:

   for i in range(1, 3):
       for j in range(10):
           for k in range(10):
               if k != j:
                   f[i][j] = min(f[i][j], f[i - 1][k] + 3 - cnt[j])

   Here, cnt[3] = 3 for column 1, and similar calculations are made.

4. Compute the Final Result:

   After processing all columns, the minimum value in the last column gives the minimum operations needed:

   result = min(f[2])

   Let’s assume after computation, result = 0, indicating that grid configurations already satisfy the conditions without further operations (this would change based on grid initializations).

This walkthrough demonstrates the dynamic programming approach to achieve the desired grid configuration, minimizing operation count while maintaining necessary distinctness between rows and columns.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n * (m + C^2)). This is because for each column i, the code calculates the frequency count of each of the 10 possible numbers in O(m) time. It then performs operations involving iterating over each pair of possible numbers j and k, which takes O(C^2) time per column, resulting in a total of O(n * (m + C^2)). Here, m is the number of rows, n is the number of columns, and C is the number of unique numbers, which is constant (C = 10).

The space complexity is O(n * C) because a 2D list f with dimensions n x 10 is used to store intermediate results for each column and possible digit, resulting in a space requirement proportional to the number of columns times the constant C.

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