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Remove All Ones With Row and Column Flips

Medium

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Topics:

ArraysGreedy AlgorithmsBit Manipulation

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

In one operation, you can choose one row or one column and flip all the values in that row or column (i.e., changing all 0s to 1s and all 1s to 0s).

Return true if it is possible to make each row of the matrix equal, otherwise, return false.

A row is considered equal if all the elements in the row are the same value (i.e., either all 0's or all 1's).

Example 1:

Input: grid = [[0,1,0],[1,0,1],[0,1,0]]
Output: true
Explanation: One possible solution is to flip the first and second columns.

Example 2:

Input: grid = [[1,1,0],[0,0,0],[0,0,0]]
Output: false
Explanation: It is impossible to make each row of the matrix equal.

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

Remove All Ones With Row and Column Flips

Medium

Asked by:

36 views

Topics:

ArraysGreedy AlgorithmsBit Manipulation

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

In one operation, you can choose one row or one column and flip all the values in that row or column (i.e., changing all 0s to 1s and all 1s to 0s).

Return true if it is possible to make each row of the matrix equal, otherwise, return false.

A row is considered equal if all the elements in the row are the same value (i.e., either all 0's or all 1's).

Example 1:

Input: grid = [[0,1,0],[1,0,1],[0,1,0]]
Output: true
Explanation: One possible solution is to flip the first and second columns.

Example 2:

Input: grid = [[1,1,0],[0,0,0],[0,0,0]]
Output: false
Explanation: It is impossible to make each row of the matrix equal.

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 input matrix, and what are the constraints on the number of rows and columns?
Are the elements in the matrix guaranteed to be only 0 or 1, or could there be other values?
If it's impossible to remove all the ones with row and column flips, what should the function return (e.g., null, -1, an empty array)?
Can I assume that the input matrix is always rectangular (i.e., all rows have the same number of columns)?
Are row and column flips independent operations, or does the order in which I apply them matter?

Brute Force Solution

Approach

The brute force approach to this problem involves trying every possible combination of row and column flips to see if we can make all the numbers in the grid become zeros. We systematically explore each option until we find one that works, or exhaust all possibilities.

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

Imagine each row and column is a light switch, which you can either flip or leave as is.
Start by not flipping anything at all. Check if the grid now contains only zeros. If so, you're done!
Now, flip the first row. Check if the grid contains only zeros. If so, you're done!
Next, flip the second row (but keep the first row unflipped). Check if the grid contains only zeros. Continue this process with each row, one at a time.
Now try flipping two rows at a time, trying every possible pair. Check each time to see if you have a grid of all zeros.
Continue this pattern trying three rows, four rows, and so on, up to flipping all the rows.
Repeat the entire process above, but this time also try flipping combinations of columns in addition to the rows. This means you need to try every combination of rows flipped/unflipped AND columns flipped/unflipped.
If, after trying every possible combination of row and column flips, you never find a grid of all zeros, then it's impossible to achieve, and you would return that it is not possible.

Code Implementation

def remove_all_ones_brute_force(grid):
    number_of_rows = len(grid)
    number_of_columns = len(grid[0])

    for row_mask in range(2**number_of_rows):
        for column_mask in range(2**number_of_columns):
            modified_grid = []
            for row_index in range(number_of_rows):
                modified_row = []
                for column_index in range(number_of_columns):
                    value = grid[row_index][column_index]

                    #Simulate row flips
                    if (row_mask >> row_index) & 1:
                        value = 1 - value

                    #Simulate column flips
                    if (column_mask >> column_index) & 1:
                        value = 1 - value

                    modified_row.append(value)
                modified_grid.append(modified_row)

            all_zeros = True
            for row_index in range(number_of_rows):
                for column_index in range(number_of_columns):
                    if modified_grid[row_index][column_index] != 0:
                        all_zeros = False
                        break
                if not all_zeros:
                    break

            #Check if the modified grid contains only zeros.
            if all_zeros:
                return True

    # If all combinations have been tried, it is impossible to achieve.
    return False

Big(O) Analysis

Time Complexity

O(2^(m+n) * m * n) – The algorithm explores all possible combinations of row and column flips. There are m rows and n columns. For each row, we have two choices: flip it or not. Similarly, for each column, we have two choices: flip it or not. Therefore, there are 2^m possible row flip combinations and 2^n possible column flip combinations. For each combination of row and column flips, we need to check if the resulting grid contains only zeros, which takes O(m * n) time. Therefore, the overall time complexity is O(2^m * 2^n * m * n), which simplifies to O(2^(m+n) * m * n).

Space Complexity

O(N*M) – The brute force approach, as described, potentially needs to create a copy of the input grid (of size N rows and M columns) to simulate each combination of row and column flips. This temporary grid allows checking if a particular combination results in all zeros without modifying the original input. Therefore, the auxiliary space is directly proportional to the size of the grid, resulting in a space complexity of O(N*M), where N is the number of rows and M is the number of columns.

Optimal Solution

Approach

The trick to efficiently solving this problem is recognizing that you don't need to try every possible flip. The matrix is essentially 'good' if all rows are either identical or perfectly opposite. We can determine if a matrix can be made 'good' by comparing each row to the first.

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

Start by picking the very first row of the grid. This will be our reference row.
Now, go through each of the other rows, one by one.
For each row, compare it to the reference row. There are two possibilities: either the row is the same as the reference row, or it's completely the opposite.
If the row is the same, great! Move on to the next row.
If the row is the opposite, we can 'flip' the entire row (change all 0s to 1s, and all 1s to 0s).
After potentially flipping the row, check if all columns are either all 0s or all 1s (or can be made that way with column flips). If you find a column that can't be made all 0s or all 1s, the grid can't be made 'good'.
If all rows can be made the same or opposite to the reference row, and all columns can be made all 0s or all 1s, then the original grid can be made 'good'. Otherwise it cannot.

Code Implementation

def solve():
    grid_matrix = [[int(num) for num in row.split()] for row in input().split(';')]
    number_of_rows = len(grid_matrix)
    number_of_columns = len(grid_matrix[0])

    # Use the first row as the reference row.
    reference_row = grid_matrix[0]

    for row_index in range(1, number_of_rows):
        current_row = grid_matrix[row_index]
        is_same = True
        is_opposite = True

        for col_index in range(number_of_columns):
            if current_row[col_index] != reference_row[col_index]:
                is_same = False
            if current_row[col_index] == reference_row[col_index]:
                is_opposite = False

        # If the row is opposite, flip it.
        if is_opposite:
            for col_index in range(number_of_columns):
                current_row[col_index] = 1 - current_row[col_index]
            grid_matrix[row_index] = current_row

    # Verify that all columns are either all 0s or all 1s.
    for col_index in range(number_of_columns):
        first_value = grid_matrix[0][col_index]
        is_valid_column = True

        for row_index in range(1, number_of_rows):
            if grid_matrix[row_index][col_index] != first_value:
                is_valid_column = False
                break

        # If a column isn't all the same, the matrix can't be good.
        if not is_valid_column:
            print("false")
            return

    print("true")

# The entry point of the script
if __name__ == "__main__":
    solve()

Big(O) Analysis

Time Complexity

O(m*n) – The algorithm iterates through each of the m rows in the grid. For each row, it compares the row to the first row, which takes O(n) time, where n is the number of columns. Therefore, comparing all rows to the first row takes O(m*n) time. The subsequent implicit column checking, to verify if all columns are uniform after potential row flips, also takes O(m*n) time. Thus, the dominant factor is iterating and checking across all cells in the matrix, resulting in O(m*n) time complexity.

Space Complexity

O(1) – The algorithm described operates primarily in-place and does not create any auxiliary data structures that scale with the input size. No extra lists, hashmaps, or significant variables are used to store intermediate results during the row comparison and potential flipping process. The only variables used are for iteration and comparison, which consume a constant amount of space regardless of the dimensions of the input grid. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty matrix input	Return true if the matrix is null or empty, as there are no ones to remove.
Matrix with only one row or one column	Check if all elements in the single row/column are the same; if not, return false, otherwise return true.
Matrix with all 0s	Return true, as no flips are needed.
Matrix with all 1s	Return true, as flipping any row or column will result in all 0s.
Matrix where no solution exists (inconsistent rows/columns)	The algorithm should detect the inconsistency and return false if no combination of flips makes all entries zero.
Large matrix exceeding memory constraints	The solution should have reasonable space complexity, ideally O(1) or O(m+n), to avoid memory issues with large matrices.
Integer overflow during row/column flip operations	Ensure the code avoids unnecessary integer operations that could lead to overflow, focusing on boolean logic.
Multiple valid solutions exist (different flip combinations)	The algorithm only needs to find if any solution exists, not all possible solutions, so stop after finding one.

Null or empty matrix input

How to Handle:

Return true if the matrix is null or empty, as there are no ones to remove.

Matrix with only one row or one column

How to Handle:

Check if all elements in the single row/column are the same; if not, return false, otherwise return true.

Matrix with all 0s

How to Handle:

Return true, as no flips are needed.

Matrix with all 1s

How to Handle:

Return true, as flipping any row or column will result in all 0s.

Matrix where no solution exists (inconsistent rows/columns)

How to Handle:

The algorithm should detect the inconsistency and return false if no combination of flips makes all entries zero.

Large matrix exceeding memory constraints

How to Handle:

The solution should have reasonable space complexity, ideally O(1) or O(m+n), to avoid memory issues with large matrices.

Integer overflow during row/column flip operations

How to Handle:

Ensure the code avoids unnecessary integer operations that could lead to overflow, focusing on boolean logic.

Multiple valid solutions exist (different flip combinations)

How to Handle:

The algorithm only needs to find if *any* solution exists, not all possible solutions, so stop after finding one.

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 input matrix, and what are the constraints on the number of rows and columns?
Are the elements in the matrix guaranteed to be only 0 or 1, or could there be other values?
If it's impossible to remove all the ones with row and column flips, what should the function return (e.g., null, -1, an empty array)?
Can I assume that the input matrix is always rectangular (i.e., all rows have the same number of columns)?
Are row and column flips independent operations, or does the order in which I apply them matter?

Brute Force Solution

Approach

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

Imagine each row and column is a light switch, which you can either flip or leave as is.
Start by not flipping anything at all. Check if the grid now contains only zeros. If so, you're done!
Now, flip the first row. Check if the grid contains only zeros. If so, you're done!
Next, flip the second row (but keep the first row unflipped). Check if the grid contains only zeros. Continue this process with each row, one at a time.
Now try flipping two rows at a time, trying every possible pair. Check each time to see if you have a grid of all zeros.
Continue this pattern trying three rows, four rows, and so on, up to flipping all the rows.
Repeat the entire process above, but this time also try flipping combinations of columns in addition to the rows. This means you need to try every combination of rows flipped/unflipped AND columns flipped/unflipped.
If, after trying every possible combination of row and column flips, you never find a grid of all zeros, then it's impossible to achieve, and you would return that it is not possible.

Code Implementation

def remove_all_ones_brute_force(grid):
    number_of_rows = len(grid)
    number_of_columns = len(grid[0])

    for row_mask in range(2**number_of_rows):
        for column_mask in range(2**number_of_columns):
            modified_grid = []
            for row_index in range(number_of_rows):
                modified_row = []
                for column_index in range(number_of_columns):
                    value = grid[row_index][column_index]

                    #Simulate row flips
                    if (row_mask >> row_index) & 1:
                        value = 1 - value

                    #Simulate column flips
                    if (column_mask >> column_index) & 1:
                        value = 1 - value

                    modified_row.append(value)
                modified_grid.append(modified_row)

            all_zeros = True
            for row_index in range(number_of_rows):
                for column_index in range(number_of_columns):
                    if modified_grid[row_index][column_index] != 0:
                        all_zeros = False
                        break
                if not all_zeros:
                    break

            #Check if the modified grid contains only zeros.
            if all_zeros:
                return True

    # If all combinations have been tried, it is impossible to achieve.
    return False

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start by picking the very first row of the grid. This will be our reference row.
Now, go through each of the other rows, one by one.
For each row, compare it to the reference row. There are two possibilities: either the row is the same as the reference row, or it's completely the opposite.
If the row is the same, great! Move on to the next row.
If the row is the opposite, we can 'flip' the entire row (change all 0s to 1s, and all 1s to 0s).
After potentially flipping the row, check if all columns are either all 0s or all 1s (or can be made that way with column flips). If you find a column that can't be made all 0s or all 1s, the grid can't be made 'good'.
If all rows can be made the same or opposite to the reference row, and all columns can be made all 0s or all 1s, then the original grid can be made 'good'. Otherwise it cannot.

Code Implementation

def solve():
    grid_matrix = [[int(num) for num in row.split()] for row in input().split(';')]
    number_of_rows = len(grid_matrix)
    number_of_columns = len(grid_matrix[0])

    # Use the first row as the reference row.
    reference_row = grid_matrix[0]

    for row_index in range(1, number_of_rows):
        current_row = grid_matrix[row_index]
        is_same = True
        is_opposite = True

        for col_index in range(number_of_columns):
            if current_row[col_index] != reference_row[col_index]:
                is_same = False
            if current_row[col_index] == reference_row[col_index]:
                is_opposite = False

        # If the row is opposite, flip it.
        if is_opposite:
            for col_index in range(number_of_columns):
                current_row[col_index] = 1 - current_row[col_index]
            grid_matrix[row_index] = current_row

    # Verify that all columns are either all 0s or all 1s.
    for col_index in range(number_of_columns):
        first_value = grid_matrix[0][col_index]
        is_valid_column = True

        for row_index in range(1, number_of_rows):
            if grid_matrix[row_index][col_index] != first_value:
                is_valid_column = False
                break

        # If a column isn't all the same, the matrix can't be good.
        if not is_valid_column:
            print("false")
            return

    print("true")

# The entry point of the script
if __name__ == "__main__":
    solve()

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty matrix input	Return true if the matrix is null or empty, as there are no ones to remove.
Matrix with only one row or one column	Check if all elements in the single row/column are the same; if not, return false, otherwise return true.
Matrix with all 0s	Return true, as no flips are needed.
Matrix with all 1s	Return true, as flipping any row or column will result in all 0s.
Matrix where no solution exists (inconsistent rows/columns)	The algorithm should detect the inconsistency and return false if no combination of flips makes all entries zero.
Large matrix exceeding memory constraints	The solution should have reasonable space complexity, ideally O(1) or O(m+n), to avoid memory issues with large matrices.
Integer overflow during row/column flip operations	Ensure the code avoids unnecessary integer operations that could lead to overflow, focusing on boolean logic.
Multiple valid solutions exist (different flip combinations)	The algorithm only needs to find if any solution exists, not all possible solutions, so stop after finding one.

Null or empty matrix input

How to Handle:

Return true if the matrix is null or empty, as there are no ones to remove.

Matrix with only one row or one column

How to Handle:

Check if all elements in the single row/column are the same; if not, return false, otherwise return true.

Matrix with all 0s

How to Handle:

Return true, as no flips are needed.

Matrix with all 1s

How to Handle:

Return true, as flipping any row or column will result in all 0s.

Matrix where no solution exists (inconsistent rows/columns)

How to Handle:

The algorithm should detect the inconsistency and return false if no combination of flips makes all entries zero.

Large matrix exceeding memory constraints

How to Handle:

The solution should have reasonable space complexity, ideally O(1) or O(m+n), to avoid memory issues with large matrices.

Integer overflow during row/column flip operations

How to Handle:

Ensure the code avoids unnecessary integer operations that could lead to overflow, focusing on boolean logic.

Multiple valid solutions exist (different flip combinations)

How to Handle:

The algorithm only needs to find if *any* solution exists, not all possible solutions, so stop after finding one.