Jobs

Interviews

Discussions

Minimum Number of Flips to Convert Binary Matrix to Zero Matrix

Hard

Asked by:

57 views

Topics:

Bit Manipulation

Given a m x n binary matrix mat. In one step, you can choose one cell and flip it and all the four neighbors of it if they exist (Flip is changing 1 to 0 and 0 to 1). A pair of cells are called neighbors if they share one edge.

Return the minimum number of steps required to convert mat to a zero matrix or -1 if you cannot.

A binary matrix is a matrix with all cells equal to 0 or 1 only.

A zero matrix is a matrix with all cells equal to 0.

Example 1:

Input: mat = [[0,0],[0,1]]
Output: 3
Explanation: One possible solution is to flip (1, 0) then (0, 1) and finally (1, 1) as shown.

Example 2:

Input: mat = [[0]]
Output: 0
Explanation: Given matrix is a zero matrix. We do not need to change it.

Example 3:

Input: mat = [[1,0,0],[1,0,0]]
Output: -1
Explanation: Given matrix cannot be a zero matrix.

Constraints:

m == mat.length
n == mat[i].length
1 <= m, n <= 3
mat[i][j] is either 0 or 1.

Try coding it in LeetCode

Minimum Number of Flips to Convert Binary Matrix to Zero Matrix

Hard

Asked by:

57 views

Topics:

Bit Manipulation

Return the minimum number of steps required to convert mat to a zero matrix or -1 if you cannot.

A binary matrix is a matrix with all cells equal to 0 or 1 only.

A zero matrix is a matrix with all cells equal to 0.

Example 1:

Input: mat = [[0,0],[0,1]]
Output: 3
Explanation: One possible solution is to flip (1, 0) then (0, 1) and finally (1, 1) as shown.

Example 2:

Input: mat = [[0]]
Output: 0
Explanation: Given matrix is a zero matrix. We do not need to change it.

Example 3:

Input: mat = [[1,0,0],[1,0,0]]
Output: -1
Explanation: Given matrix cannot be a zero matrix.

Constraints:

m == mat.length
n == mat[i].length
1 <= m, n <= 3
mat[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 (number of rows and columns) of the binary matrix, and what is the maximum size I should expect?
Is the input matrix guaranteed to be rectangular (i.e., each row has the same number of columns)?
If it's impossible to convert the binary matrix to a zero matrix through any number of flips, what should the function return?
When flipping a cell, should I also flip its diagonal neighbors, or only the immediate horizontal and vertical neighbors?
Is the original input matrix allowed to be modified, or should I work on a copy?

Brute Force Solution

Approach

The brute force strategy tries every single possible combination of flips to the matrix. We try flipping different cells one at a time, then two at a time, and so on. Eventually, we find the sequence of flips that turns the entire matrix into zeros, and we keep track of the fewest number of flips needed.

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

First, consider the possibility of not flipping any cell in the matrix at all. Check if the matrix is already all zeros. If it is, then you are done and the answer is zero.
Next, consider flipping only one cell at a time. For each cell in the matrix, flip it, and then check if the entire matrix is now all zeros. If it is, record that we found a solution with one flip.
Then, consider flipping two cells at a time. Try every possible pair of cells, flip both of them, and check if the entire matrix is all zeros. If it is, record that we found a solution with two flips.
Continue this process, trying all possible combinations of three flips, four flips, and so on, flipping every possible subset of cells and checking if it results in an all-zero matrix.
As you find solutions, keep track of the minimum number of flips it took to achieve an all-zero matrix.
Once you've explored all possible combinations of flips (up to the point where the number of flips equals the total number of cells in the matrix), the smallest number you recorded is the answer.

Code Implementation

def min_flips_brute_force(matrix):
    rows = len(matrix)
    cols = len(matrix[0])
    total_cells = rows * cols
    min_flips = float('inf')

    for num_flips in range(total_cells + 1):
        for combination in get_combinations(total_cells, num_flips):
            temp_matrix = [row[:] for row in matrix]

            # Apply the current combination of flips to a temporary matrix
            for index in combination:
                row_index = index // cols
                col_index = index % cols
                flip_cell_and_neighbors(temp_matrix, row_index, col_index)

            # Check if the temporary matrix is all zeros after flips
            if is_zero_matrix(temp_matrix):

                #Keep track of minimum flips to reach zero matrix
                min_flips = min(min_flips, num_flips)

    if min_flips == float('inf'):
        return -1
    return min_flips

def get_combinations(total_cells, num_flips):
    if num_flips == 0:
        yield []
        return
    if num_flips > total_cells:
        return

    def generate_combinations(start_index, current_combination):
        if len(current_combination) == num_flips:
            yield current_combination[:]
            return
        
        for i in range(start_index, total_cells):
            current_combination.append(i)
            yield from generate_combinations(i + 1, current_combination)
            current_combination.pop()

    yield from generate_combinations(0, [])

def flip_cell_and_neighbors(matrix, row_index, col_index):
    rows = len(matrix)
    cols = len(matrix[0])

    def flip(row_index, col_index):
        if 0 <= row_index < rows and 0 <= col_index < cols:
            matrix[row_index][col_index] = 1 - matrix[row_index][col_index]

    flip(row_index, col_index)
    flip(row_index - 1, col_index)
    flip(row_index + 1, col_index)
    flip(row_index, col_index - 1)
    flip(row_index, col_index + 1)

def is_zero_matrix(matrix):

    #Check if matrix contains only zeros
    for row in matrix:
        for cell in row:
            if cell != 0:
                return False
    return True

Big(O) Analysis

Time Complexity

O(2^(m*n)) – The algorithm iterates through all possible combinations of cell flips in the matrix. If the matrix has m rows and n columns, there are a total of m*n cells. Each cell can either be flipped or not flipped, leading to 2^(m*n) possible combinations. For each combination, the algorithm needs to check if the resulting matrix is a zero matrix, which takes O(m*n) time. Therefore, the overall time complexity is O(m*n * 2^(m*n)). While checking if the matrix is all zeroes takes O(m*n), the exponential component dominates. Thus, simplifying by removing the m*n check within the loop, the time complexity is approximately O(2^(m*n)).

Space Complexity

O(1) – The described brute force approach primarily involves flipping cells in the matrix and checking for an all-zero state. It doesn't explicitly mention creating new data structures that scale with the input size (N), where N represents the number of cells in the matrix. The algorithm keeps track of the minimum number of flips, which uses constant space. Therefore, the auxiliary space complexity is constant, or O(1).

Optimal Solution

Approach

The key idea is to realize that flipping a particular cell affects its neighbors in a predictable way. Because of this, we only need to consider each cell as a possible 'first flip' and then let that choice determine all subsequent flips. This drastically reduces the amount of searching we need to do.

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

Consider the first cell in the matrix. We have two choices: either flip it or don't flip it.
For each of these choices, proceed to the next cell in a left-to-right, top-to-bottom order.
The decision to flip each subsequent cell is now forced by the previous cells. We must flip the current cell if its 'up' neighbor is not zero.
Continue this process of forced flips until you reach the end of the matrix.
Check if the final matrix is all zeros. If it is, keep track of how many flips were needed.
Repeat the entire process starting with the other choice for the first cell (if you flipped it before, don't flip it now, and vice versa).
After exploring both initial choices, select the flip combination that resulted in a zero matrix with the fewest total flips. If neither results in a zero matrix, it's impossible.

Code Implementation

def min_flips(matrix):
    rows = len(matrix)
    cols = len(matrix[0])
    min_flips_needed = float('inf')

    def flip_cell(current_matrix, row, col):
        if 0 <= row < rows and 0 <= col < cols:
            current_matrix[row][col] ^= 1

    def calculate_flips(initial_flip):
        current_matrix = [row[:] for row in matrix]
        flips = 0
        if initial_flip:
            flip_cell(current_matrix, 0, 0)
            flips += 1

        # We iterate through the matrix
        for row in range(rows):
            for col in range(cols):
                if row > 0 and current_matrix[row - 1][col] == 1:
                    # Flip the current cell if the top neighbor is 1
                    flip_cell(current_matrix, row, col)
                    flip_cell(current_matrix, row - 1, col)
                    flips += 1
                    if col > 0:
                        flip_cell(current_matrix, row, col - 1)
                    if col < cols - 1:
                        flip_cell(current_matrix, row, col + 1)
                    if row < rows - 1:
                        flip_cell(current_matrix, row + 1, col)

        is_zero_matrix = all(current_matrix[row][col] == 0 for row in range(rows) for col in range(cols))
        if is_zero_matrix:
            return flips
        else:
            return float('inf')

    # Consider both initial states:
    min_flips_needed = min(min_flips_needed, calculate_flips(True), calculate_flips(False))

    if min_flips_needed == float('inf'):
        return -1
    else:
        return min_flips_needed

Big(O) Analysis

Time Complexity

O(m*n*2^(m*n)) – Let m be the number of rows and n be the number of columns in the binary matrix. The described approach explores two possibilities (flip or don't flip) for the first cell. This initial choice cascades, forcing decisions for subsequent cells. We are essentially iterating through 2^(m*n) possibilities of starting states of the first cell. For each of these possibilities, we are iterating through each of the m*n cells to decide if we flip based on the adjacent cell. Then, we check each of the m*n cells to see if it is zero. Thus, the Big O complexity is approximately (m*n) * 2^(m*n) * (m*n), which simplifies to O(m*n*2^(m*n)) since we are only concerned with whether it results in a zero matrix.

Space Complexity

O(N) – The space complexity is determined by the need to create a copy of the input matrix. The algorithm explores different flipping scenarios and modifies a copy of the original matrix to test these scenarios. This copy has the same dimensions as the input matrix. Therefore, the auxiliary space required is proportional to the number of elements in the matrix, which we can denote as N. Thus, the space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty matrix	Return 0 if the matrix is null or has zero rows or columns, as it's already a zero matrix.
1x1 matrix with a 1	Return 1, as a single flip makes it a zero matrix.
All cells are already 0	Return 0, as no flips are needed.
All cells are 1	The number of flips is related to the problem's definition of flipping a cell and its neighbors, so handle this by finding the minimum flips from each starting cell.
Matrix with only one row or one column	Check the definition of neighbors (up, down, left, right) when a cell is on a boundary, implementing special boundary checks.
Large matrix that might cause a stack overflow with recursion	Use an iterative BFS approach instead of a recursive DFS to avoid stack overflow issues.
No solution exists (impossible to convert)	If the BFS queue empties without finding the zero matrix, return -1 to indicate no solution is possible.
Integer overflow when calculating matrix state	Use bit manipulation or a suitable data type (e.g., long) to represent the matrix state and prevent overflow.

Null or empty matrix

How to Handle:

Return 0 if the matrix is null or has zero rows or columns, as it's already a zero matrix.

1x1 matrix with a 1

How to Handle:

Return 1, as a single flip makes it a zero matrix.

All cells are already 0

How to Handle:

Return 0, as no flips are needed.

All cells are 1

How to Handle:

The number of flips is related to the problem's definition of flipping a cell and its neighbors, so handle this by finding the minimum flips from each starting cell.

Matrix with only one row or one column

How to Handle:

Check the definition of neighbors (up, down, left, right) when a cell is on a boundary, implementing special boundary checks.

Large matrix that might cause a stack overflow with recursion

How to Handle:

Use an iterative BFS approach instead of a recursive DFS to avoid stack overflow issues.

No solution exists (impossible to convert)

How to Handle:

If the BFS queue empties without finding the zero matrix, return -1 to indicate no solution is possible.

Integer overflow when calculating matrix state

How to Handle:

Use bit manipulation or a suitable data type (e.g., long) to represent the matrix state and prevent overflow.

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 (number of rows and columns) of the binary matrix, and what is the maximum size I should expect?
Is the input matrix guaranteed to be rectangular (i.e., each row has the same number of columns)?
If it's impossible to convert the binary matrix to a zero matrix through any number of flips, what should the function return?
When flipping a cell, should I also flip its diagonal neighbors, or only the immediate horizontal and vertical neighbors?
Is the original input matrix allowed to be modified, or should I work on a copy?

Brute Force Solution

Approach

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

First, consider the possibility of not flipping any cell in the matrix at all. Check if the matrix is already all zeros. If it is, then you are done and the answer is zero.
Next, consider flipping only one cell at a time. For each cell in the matrix, flip it, and then check if the entire matrix is now all zeros. If it is, record that we found a solution with one flip.
Then, consider flipping two cells at a time. Try every possible pair of cells, flip both of them, and check if the entire matrix is all zeros. If it is, record that we found a solution with two flips.
Continue this process, trying all possible combinations of three flips, four flips, and so on, flipping every possible subset of cells and checking if it results in an all-zero matrix.
As you find solutions, keep track of the minimum number of flips it took to achieve an all-zero matrix.
Once you've explored all possible combinations of flips (up to the point where the number of flips equals the total number of cells in the matrix), the smallest number you recorded is the answer.

Code Implementation

def min_flips_brute_force(matrix):
    rows = len(matrix)
    cols = len(matrix[0])
    total_cells = rows * cols
    min_flips = float('inf')

    for num_flips in range(total_cells + 1):
        for combination in get_combinations(total_cells, num_flips):
            temp_matrix = [row[:] for row in matrix]

            # Apply the current combination of flips to a temporary matrix
            for index in combination:
                row_index = index // cols
                col_index = index % cols
                flip_cell_and_neighbors(temp_matrix, row_index, col_index)

            # Check if the temporary matrix is all zeros after flips
            if is_zero_matrix(temp_matrix):

                #Keep track of minimum flips to reach zero matrix
                min_flips = min(min_flips, num_flips)

    if min_flips == float('inf'):
        return -1
    return min_flips

def get_combinations(total_cells, num_flips):
    if num_flips == 0:
        yield []
        return
    if num_flips > total_cells:
        return

    def generate_combinations(start_index, current_combination):
        if len(current_combination) == num_flips:
            yield current_combination[:]
            return
        
        for i in range(start_index, total_cells):
            current_combination.append(i)
            yield from generate_combinations(i + 1, current_combination)
            current_combination.pop()

    yield from generate_combinations(0, [])

def flip_cell_and_neighbors(matrix, row_index, col_index):
    rows = len(matrix)
    cols = len(matrix[0])

    def flip(row_index, col_index):
        if 0 <= row_index < rows and 0 <= col_index < cols:
            matrix[row_index][col_index] = 1 - matrix[row_index][col_index]

    flip(row_index, col_index)
    flip(row_index - 1, col_index)
    flip(row_index + 1, col_index)
    flip(row_index, col_index - 1)
    flip(row_index, col_index + 1)

def is_zero_matrix(matrix):

    #Check if matrix contains only zeros
    for row in matrix:
        for cell in row:
            if cell != 0:
                return False
    return True

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Consider the first cell in the matrix. We have two choices: either flip it or don't flip it.
For each of these choices, proceed to the next cell in a left-to-right, top-to-bottom order.
The decision to flip each subsequent cell is now forced by the previous cells. We must flip the current cell if its 'up' neighbor is not zero.
Continue this process of forced flips until you reach the end of the matrix.
Check if the final matrix is all zeros. If it is, keep track of how many flips were needed.
Repeat the entire process starting with the other choice for the first cell (if you flipped it before, don't flip it now, and vice versa).
After exploring both initial choices, select the flip combination that resulted in a zero matrix with the fewest total flips. If neither results in a zero matrix, it's impossible.

Code Implementation

def min_flips(matrix):
    rows = len(matrix)
    cols = len(matrix[0])
    min_flips_needed = float('inf')

    def flip_cell(current_matrix, row, col):
        if 0 <= row < rows and 0 <= col < cols:
            current_matrix[row][col] ^= 1

    def calculate_flips(initial_flip):
        current_matrix = [row[:] for row in matrix]
        flips = 0
        if initial_flip:
            flip_cell(current_matrix, 0, 0)
            flips += 1

        # We iterate through the matrix
        for row in range(rows):
            for col in range(cols):
                if row > 0 and current_matrix[row - 1][col] == 1:
                    # Flip the current cell if the top neighbor is 1
                    flip_cell(current_matrix, row, col)
                    flip_cell(current_matrix, row - 1, col)
                    flips += 1
                    if col > 0:
                        flip_cell(current_matrix, row, col - 1)
                    if col < cols - 1:
                        flip_cell(current_matrix, row, col + 1)
                    if row < rows - 1:
                        flip_cell(current_matrix, row + 1, col)

        is_zero_matrix = all(current_matrix[row][col] == 0 for row in range(rows) for col in range(cols))
        if is_zero_matrix:
            return flips
        else:
            return float('inf')

    # Consider both initial states:
    min_flips_needed = min(min_flips_needed, calculate_flips(True), calculate_flips(False))

    if min_flips_needed == float('inf'):
        return -1
    else:
        return min_flips_needed

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty matrix	Return 0 if the matrix is null or has zero rows or columns, as it's already a zero matrix.
1x1 matrix with a 1	Return 1, as a single flip makes it a zero matrix.
All cells are already 0	Return 0, as no flips are needed.
All cells are 1	The number of flips is related to the problem's definition of flipping a cell and its neighbors, so handle this by finding the minimum flips from each starting cell.
Matrix with only one row or one column	Check the definition of neighbors (up, down, left, right) when a cell is on a boundary, implementing special boundary checks.
Large matrix that might cause a stack overflow with recursion	Use an iterative BFS approach instead of a recursive DFS to avoid stack overflow issues.
No solution exists (impossible to convert)	If the BFS queue empties without finding the zero matrix, return -1 to indicate no solution is possible.
Integer overflow when calculating matrix state	Use bit manipulation or a suitable data type (e.g., long) to represent the matrix state and prevent overflow.

Null or empty matrix

How to Handle:

Return 0 if the matrix is null or has zero rows or columns, as it's already a zero matrix.

1x1 matrix with a 1

How to Handle:

Return 1, as a single flip makes it a zero matrix.

All cells are already 0

How to Handle:

Return 0, as no flips are needed.

All cells are 1

How to Handle:

The number of flips is related to the problem's definition of flipping a cell and its neighbors, so handle this by finding the minimum flips from each starting cell.

Matrix with only one row or one column

How to Handle:

Check the definition of neighbors (up, down, left, right) when a cell is on a boundary, implementing special boundary checks.

Large matrix that might cause a stack overflow with recursion

How to Handle:

Use an iterative BFS approach instead of a recursive DFS to avoid stack overflow issues.

No solution exists (impossible to convert)

How to Handle:

If the BFS queue empties without finding the zero matrix, return -1 to indicate no solution is possible.

Integer overflow when calculating matrix state

How to Handle:

Use bit manipulation or a suitable data type (e.g., long) to represent the matrix state and prevent overflow.