Minimum Operations to Remove Adjacent Ones in Matrix

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Dynamic ProgrammingBit Manipulation

You are given a 0-indexed m x n binary matrix grid.

In one operation, you can choose a cell (i, j) such that grid[i][j] == 1 and change it to 0. However, if a cell (i, j) has two adjacent cells with value 1, you are not allowed to change it.

Return the minimum number of operations needed to remove all 1s from grid.

Example 1:

Input: grid = [[0,1,0,0],[1,0,0,1],[0,1,0,0]]
Output: 0
Explanation: There are no cells with two adjacent cells with value 1 so no operations are needed.

Example 2:

Input: grid = [[1,1,0,0],[0,1,1,0],[0,0,1,1]]
Output: 3
Explanation: You can do the following operations:
- Choose cell (0,0). The grid becomes [[0,1,0,0],[0,1,1,0],[0,0,1,1]].
- Choose cell (0,1). The grid becomes [[0,0,0,0],[0,1,1,0],[0,0,1,1]].
- Choose cell (2,3). The grid becomes [[0,0,0,0],[0,1,1,0],[0,0,1,0]].
There are no more possible operations so the answer is 3.

Example 3:

Input: grid = [[1,1,1],[1,1,1]]
Output: 2
Explanation: You can do the following operations:
- Choose cell (0,1). The grid becomes [[1,0,1],[1,1,1]].
- Choose cell (1,1). The grid becomes [[1,0,1],[1,0,1]].
There are no more possible operations so the answer is 2.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] is either 0 or 1.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the dimensions of the matrix, and what is the maximum possible size of the matrix (number of rows and columns)?
Can the matrix contain values other than 0 and 1, or is it guaranteed to only contain binary values?
If there are no adjacent 1s in the matrix, should I return 0 or some other specific value?
By 'adjacent', do you mean only horizontally or vertically adjacent, or also diagonally adjacent?
Is the matrix guaranteed to be rectangular (i.e., all rows have the same number of columns)? If not, what should I do if rows have differing lengths?

Brute Force Solution

Approach

The brute force approach to this problem is like trying every single possible combination of operations to remove adjacent ones and see which one takes the fewest steps. It's inefficient, but it guarantees we'll find the absolute best answer.

Here's how the algorithm would work step-by-step:

Consider the matrix as it is now. This is our starting point.
Find every pair of adjacent ones in the matrix. Each pair represents a possible operation we could perform.
For each pair of adjacent ones, imagine we perform that operation, which removes those ones. This creates a new matrix.
For each new matrix we've created, again find every pair of adjacent ones and imagine performing each of those operations, creating even more matrices.
Keep doing this, branching out like a tree, until all possible sequences of operations have been tried and there are no adjacent ones left in any of the matrices.
For each path we took (each sequence of operations), count how many operations it took to get to a matrix with no adjacent ones.
Finally, compare all the counts and choose the smallest one. That smallest count is the minimum number of operations required.

Code Implementation

def minimum_operations_to_remove_adjacent_ones_brute_force(matrix):
    rows = len(matrix)
    cols = len(matrix[0]) if rows > 0 else 0

    def find_adjacent_ones(current_matrix):
        adjacent_ones_pairs = []
        for row_index in range(rows):
            for col_index in range(cols):
                if current_matrix[row_index][col_index] == 1:
                    # Check right
                    if col_index + 1 < cols and current_matrix[row_index][col_index + 1] == 1:
                        adjacent_ones_pairs.append(((row_index, col_index), (row_index, col_index + 1)))
                    # Check down
                    if row_index + 1 < rows and current_matrix[row_index + 1][col_index] == 1:
                        adjacent_ones_pairs.append(((row_index, col_index), (row_index + 1, col_index)))
        return adjacent_ones_pairs

    def apply_operation(current_matrix, pair):
        new_matrix = [row[:] for row in current_matrix]
        (row1, col1), (row2, col2) = pair
        new_matrix[row1][col1] = 0
        new_matrix[row2][col2] = 0
        return new_matrix

    min_operations = float('inf')

    def solve(current_matrix, operations_count):
        nonlocal min_operations
        adjacent_ones_pairs = find_adjacent_ones(current_matrix)

        #If no adjacent ones, update min operations if necessary
        if not adjacent_ones_pairs:
            min_operations = min(min_operations, operations_count)
            return

        # Try removing each pair of adjacent ones
        for pair in adjacent_ones_pairs:
            new_matrix = apply_operation(current_matrix, pair)

            #Recursively explore the next possible states
            solve(new_matrix, operations_count + 1)
   solve(matrix, 0)
   # Account for the possibility that the initial state has no adjacent ones
    if min_operations == float('inf'):
        if not find_adjacent_ones(matrix):
            return 0
   return min_operations

Big(O) Analysis

Time Complexity

O(2^(m*n)) – In the worst-case scenario, nearly every possible combination of adjacent ones needs to be evaluated. With an m x n matrix, there can be a significant number of adjacent pairs. Each identified adjacent pair leads to a new branch of possibilities. Because we are trying every possible combination, the number of branches grows exponentially. The maximum number of steps will be proportional to 2 to the power of the number of cells (m*n), giving a time complexity of O(2^(m*n)).

Space Complexity

O(2^N) – The brute force approach explores all possible combinations of removing adjacent ones. Each operation creates a new matrix, leading to a branching tree-like structure. In the worst-case scenario, where N represents the number of cells in the matrix (rows * columns), each adjacent pair removal could potentially generate a unique matrix. Consequently, the space complexity grows exponentially as we store these intermediate matrices for further exploration, leading to a space usage proportional to O(2^N) where N is the number of cells. The algorithm implicitly maintains a queue or stack (through recursion) to manage the exploration of these matrices.

Optimal Solution

Approach

We need to eliminate adjacent '1's in the matrix with the fewest steps. The key is to use dynamic programming to consider each cell and the best possible operation to remove those pairs of ones efficiently, building up the solution from smaller subproblems.

Here's how the algorithm would work step-by-step:

Think of each cell in the matrix. We'll calculate the minimum number of operations to make the area above and to the left of that cell have no adjacent ones.
For each cell, consider two options: either we flip the current cell's value, or we don't.
If we flip the current cell, we add one operation to the running total and then consider the impact this has on adjacent cells.
If we don't flip the cell, we need to make sure that its neighbors (above and to the left) are either 0 or have already been handled.
To make the best decision, we compare the number of operations required in both the flip and no-flip scenarios.
We store the minimum number of operations needed up to each cell in a separate data structure to avoid recalculating it later.
The final result is the minimum number of operations needed for the entire matrix, which is the value stored for the last cell we processed.

Code Implementation

def min_operations_to_remove_adjacent_ones(matrix):
    rows = len(matrix)
    cols = len(matrix[0])

    # dp[i][j] stores the minimum operations to make the submatrix
    #  above and to the left of (i, j) have no adjacent ones.
    dp = [[0] * cols for _ in range(rows)]

    for row_index in range(rows):
        for col_index in range(cols):
            # Consider flipping the current cell.
            flip_operations = 1
            if row_index > 0:
                flip_operations += dp[row_index - 1][col_index]
            if col_index > 0:
                flip_operations += dp[row_index][col_index - 1]
            if row_index > 0 and col_index > 0:
                flip_operations -= dp[row_index - 1][col_index - 1]

            # Consider not flipping the current cell.
            no_flip_operations = float('inf')
            is_valid_without_flip = True

            if row_index > 0 and matrix[row_index - 1][col_index] == matrix[row_index][col_index] == 1:
                is_valid_without_flip = False
            if col_index > 0 and matrix[row_index][col_index - 1] == matrix[row_index][col_index] == 1:
                is_valid_without_flip = False

            if is_valid_without_flip:
                no_flip_operations = 0
                if row_index > 0:
                    no_flip_operations += dp[row_index - 1][col_index]
                if col_index > 0:
                    no_flip_operations += dp[row_index][col_index - 1]
                if row_index > 0 and col_index > 0:
                    no_flip_operations -= dp[row_index - 1][col_index - 1]
            
            # We compare flipping vs not flipping to get the minimum operations.
            dp[row_index][col_index] = min(flip_operations, no_flip_operations)

    # This is where we get our final answer.
    return dp[rows - 1][cols - 1]

Big(O) Analysis

Time Complexity

O(m*n) – The dynamic programming approach iterates through each cell of the m x n matrix. For each cell, a constant amount of work is performed, involving checking adjacent cells (above and to the left) and comparing the 'flip' and 'no-flip' scenarios to determine the minimum operations. Since we visit each cell once and perform a fixed number of operations per cell, the overall time complexity is proportional to the number of cells in the matrix, which is m*n, leading to O(m*n).

Space Complexity

O(N*M) – The algorithm uses a separate data structure to store the minimum number of operations needed up to each cell. Since the input matrix is of size N*M (where N is the number of rows and M is the number of columns), this data structure, which likely resembles a DP table or similar, will also be of size N*M. Therefore, the auxiliary space used is directly proportional to the size of the input matrix, resulting in a space complexity of O(N*M).

Edge Cases

Case	How to Handle
Null or empty matrix	Return 0 immediately, as no operations are needed on an empty matrix.
Matrix with a single row or column	Iterate through the row or column and count adjacent 1s.
Matrix filled entirely with zeros	Return 0 immediately, as no adjacent 1s exist.
Matrix filled entirely with ones	The number of operations required is (rows * cols) - 1, or fewer if a more efficient removal strategy is implemented, but ensure the logic handles this dense case correctly to avoid stack overflow in recursive solutions or incorrect operation count.
Large matrix dimensions (potential for stack overflow with recursive DFS/BFS)	Implement iterative DFS/BFS instead of recursive to avoid stack overflow errors.
Disconnected groups of adjacent ones	Ensure the algorithm correctly identifies and processes all disconnected groups of ones, summing the required operations for each group.
Integer overflow when calculating the number of operations	Use a data type (e.g., long in Java) that can accommodate the maximum possible number of operations.
Matrix with dimensions near integer limits	Verify the calculations of indices and sizes don't overflow and cause unexpected behavior.

Null or empty matrix

How to Handle:

Return 0 immediately, as no operations are needed on an empty matrix.

Matrix with a single row or column

How to Handle:

Iterate through the row or column and count adjacent 1s.

Matrix filled entirely with zeros

How to Handle:

Return 0 immediately, as no adjacent 1s exist.

Matrix filled entirely with ones

How to Handle:

The number of operations required is (rows * cols) - 1, or fewer if a more efficient removal strategy is implemented, but ensure the logic handles this dense case correctly to avoid stack overflow in recursive solutions or incorrect operation count.

Large matrix dimensions (potential for stack overflow with recursive DFS/BFS)

How to Handle:

Implement iterative DFS/BFS instead of recursive to avoid stack overflow errors.

Disconnected groups of adjacent ones

How to Handle:

Ensure the algorithm correctly identifies and processes all disconnected groups of ones, summing the required operations for each group.

Integer overflow when calculating the number of operations

How to Handle:

Use a data type (e.g., long in Java) that can accommodate the maximum possible number of operations.

Matrix with dimensions near integer limits

How to Handle:

Verify the calculations of indices and sizes don't overflow and cause unexpected behavior.