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Remove All Ones With Row and Column Flips II

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Topics:

ArraysRecursionDynamic Programming

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., change all 0s to 1s, and all 1s to 0s).

Return the maximum number of 1s you can have in the matrix after performing an arbitrary number of operations.

Example 1:

Input: grid = [[0,0,1],[1,1,0],[1,0,0]]
Output: 6
Explanation: We can choose to flip the 0^th column and the 2^nd column.

The final number of 1s is 6.

Example 2:

Input: grid = [[0]]
Output: 1
Explanation: We can choose to flip the 0^th row.
The final number of 1s is 1.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 15
grid[i][j] is either 0 or 1.

Try coding it in LeetCode

Remove All Ones With Row and Column Flips II

Medium

Asked by:

26 views

Topics:

ArraysRecursionDynamic Programming

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., change all 0s to 1s, and all 1s to 0s).

Return the maximum number of 1s you can have in the matrix after performing an arbitrary number of operations.

Example 1:

Input: grid = [[0,0,1],[1,1,0],[1,0,0]]
Output: 6
Explanation: We can choose to flip the 0^th column and the 2^nd column.

The final number of 1s is 6.

Example 2:

Input: grid = [[0]]
Output: 1
Explanation: We can choose to flip the 0^th row.
The final number of 1s is 1.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 15
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 are the constraints on the dimensions (e.g., maximum number of rows and columns)?
What are the possible values within the matrix? Are they guaranteed to be only 0 or 1?
If it's not possible to remove all the 1s, what should I return? Should I return `false`, or throw an exception, or something else?
Is it guaranteed that the input matrix is rectangular (i.e., all rows have the same number of columns)?
Could you clarify what 'flipping a row/column' precisely means? Does it change 0s to 1s and 1s to 0s, or does it perform another operation?

Brute Force Solution

Approach

The brute force approach tries every possible combination of flips to see if we can make the entire grid contain only zeros. We systematically go through each possible setting of rows and columns.

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

Consider the possibility of flipping no rows or columns.
Then, consider flipping only the first row, or only the second row, and so on, trying each row individually.
Next, consider flipping every pair of rows, then every group of three rows, and so on, until you've considered flipping all possible combinations of rows.
For each of these row-flipping scenarios, repeat the same process for the columns. Try flipping no columns, then each column individually, then every pair of columns, and so on, until you've tried all combinations of columns.
After flipping a specific set of rows and columns, check if the entire grid now contains only zeros.
If the grid contains only zeros, remember this combination as a valid solution.
After exploring all possible combinations of row and column flips, determine if any valid solutions were found. If a solution was found, the grid can be converted to all zeroes.

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):
            flipped_grid = [row[:] for row in grid]

            # Flip rows based on the row_mask
            for row_index in range(number_of_rows):
                if (row_mask >> row_index) & 1:
                    for column_index in range(number_of_columns):
                        flipped_grid[row_index][column_index] = 1 - flipped_grid[row_index][column_index]

            # Flip columns based on the column_mask
            for column_index in range(number_of_columns):
                if (column_mask >> column_index) & 1:
                    for row_index in range(number_of_rows):
                        flipped_grid[row_index][column_index] = 1 - flipped_grid[row_index][column_index]

            # Check if all elements are now zero
            all_zeros = True
            for row_index in range(number_of_rows):
                for column_index in range(number_of_columns):
                    if flipped_grid[row_index][column_index] == 1:
                        all_zeros = False
                        break
                if not all_zeros:
                    break

            # If a solution is found, return True
            if all_zeros:
                return True

    # If no solution found, return False
    return False

Big(O) Analysis

Time Complexity

O(2^(m+n) * m * n) – The algorithm iterates through all possible combinations of row flips, which takes O(2^m) time, where m is the number of rows. For each row combination, it iterates through all possible combinations of column flips, which takes O(2^n) time, where n is the number of columns. After each combination of row and column flips, the algorithm checks if the entire 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(1) – The provided brute force approach operates primarily by flipping the original grid in place and checking if it results in all zeros. It does not allocate significant auxiliary space for storing intermediate results or data structures beyond a few boolean variables to track if a solution is found. The space used for these variables remains constant regardless of the size of the input grid (number of rows and columns). Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The problem asks to determine if a matrix can be converted to all zeros by flipping rows and columns. The efficient approach avoids trying every flip combination and instead focuses on identifying relationships between rows. If rows are fundamentally incompatible, the task is impossible.

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

Check if any two rows are inherently contradictory. Two rows are contradictory if, no matter how columns are flipped, you can't make them the same.
Specifically, for each pair of rows, see if either the rows are identical or if they are the inverse of each other (where every cell in one row is the opposite of the other).
If any row pair is neither identical nor inverse, then there's no way to make the whole matrix all zeros by flipping rows and columns, so you can immediately say it's not possible.
If all pairs of rows are either identical or inverse, it means a sequence of row and column flips could make the whole grid zero, so the answer is yes.

Code Implementation

def solve():
    matrix = [
        [0, 1, 0],
        [1, 0, 1],
        [0, 1, 0]
    ]

    number_of_rows = len(matrix)

    # Check if all pairs of rows are identical or inverse.
    for i in range(number_of_rows):
        for j in range(i + 1, number_of_rows):
            rows_are_identical = True
            rows_are_inverse = True

            for column_index in range(len(matrix[0])):
                if matrix[i][column_index] != matrix[j][column_index]:
                    rows_are_identical = False

                if matrix[i][column_index] == matrix[j][column_index]:
                    rows_are_inverse = False

            # If the rows are neither identical nor inverse, return false
            if not rows_are_identical and not rows_are_inverse:
                return False

    # All rows are either identical or inverse.
    return True

def remove_all_ones(matrix):
    number_of_rows = len(matrix)
    if number_of_rows <= 1:
        return True

    #Iterate through all possible row pairs
    for row_index_1 in range(number_of_rows):
        for row_index_2 in range(row_index_1 + 1, number_of_rows):

            rows_are_identical = True
            rows_are_inverse = True
            number_of_columns = len(matrix[0])

            for column_index in range(number_of_columns):
                if matrix[row_index_1][column_index] != matrix[row_index_2][column_index]:
                    rows_are_identical = False

                if matrix[row_index_1][column_index] == matrix[row_index_2][column_index]:
                    rows_are_inverse = False
            
            #Important: If rows aren't identical/inverse, transformation is impossible
            if not rows_are_identical and not rows_are_inverse:
                return False

    return True

Big(O) Analysis

Time Complexity

O(m*n^2) – The algorithm iterates through all possible pairs of rows in the matrix, where 'n' represents the number of rows. For each pair of rows, it compares them to determine if they are identical or the inverse of each other. This comparison takes O(m) time, where 'm' is the number of columns. Since there are approximately n * n/2 pairs of rows, the overall time complexity is O(m * n * n/2), which simplifies to O(m*n^2).

Space Complexity

O(1) – The algorithm iterates through pairs of rows, comparing them for equality or inverse relationship. It doesn't create any auxiliary data structures whose size scales with the input matrix dimensions (rows x columns). The comparisons are done in-place or with a few constant-sized variables. Therefore, the auxiliary space used is constant, independent of the input size N (number of rows and columns), resulting in O(1) space complexity.

Edge Cases

Case	How to Handle
Null or empty matrix	Return true if the matrix is null or empty, as no '1's exist, thus satisfying the condition.
Matrix with only one row or one column	Check if a single row or column contains only 0s after potential flipping; if so, return true; otherwise, return false.
Matrix with all zeros	Return true immediately as the condition is already met (no ones to remove).
Matrix with all ones	Test all combinations of row and column flips to determine if a valid solution exists; if not, return false.
Large matrix (scalability)	Ensure the algorithm's time complexity is optimized (e.g., avoiding exhaustive search when possible) to handle large inputs within reasonable time limits.
A case where no solution exists, i.e., no combination of flips leads to all zeros	The algorithm should correctly identify when no solution is possible and return false.
Integer overflow when calculating row/column flips	Use appropriate data types (e.g., long) to prevent integer overflow when counting the number of '1's in rows and columns.
Matrix with mostly ones and very few zeros	This input can cause the algorithm to explore many invalid combinations of row/column flips, so efficient pruning/backtracking is needed to improve performance.

Null or empty matrix

How to Handle:

Return true if the matrix is null or empty, as no '1's exist, thus satisfying the condition.

Matrix with only one row or one column

How to Handle:

Check if a single row or column contains only 0s after potential flipping; if so, return true; otherwise, return false.

Matrix with all zeros

How to Handle:

Return true immediately as the condition is already met (no ones to remove).

Matrix with all ones

How to Handle:

Test all combinations of row and column flips to determine if a valid solution exists; if not, return false.

Large matrix (scalability)

How to Handle:

Ensure the algorithm's time complexity is optimized (e.g., avoiding exhaustive search when possible) to handle large inputs within reasonable time limits.

A case where no solution exists, i.e., no combination of flips leads to all zeros

How to Handle:

The algorithm should correctly identify when no solution is possible and return false.

Integer overflow when calculating row/column flips

How to Handle:

Use appropriate data types (e.g., long) to prevent integer overflow when counting the number of '1's in rows and columns.

Matrix with mostly ones and very few zeros

How to Handle:

This input can cause the algorithm to explore many invalid combinations of row/column flips, so efficient pruning/backtracking is needed to improve performance.

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 are the constraints on the dimensions (e.g., maximum number of rows and columns)?
What are the possible values within the matrix? Are they guaranteed to be only 0 or 1?
If it's not possible to remove all the 1s, what should I return? Should I return `false`, or throw an exception, or something else?
Is it guaranteed that the input matrix is rectangular (i.e., all rows have the same number of columns)?
Could you clarify what 'flipping a row/column' precisely means? Does it change 0s to 1s and 1s to 0s, or does it perform another operation?

Brute Force Solution

Approach

The brute force approach tries every possible combination of flips to see if we can make the entire grid contain only zeros. We systematically go through each possible setting of rows and columns.

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

Consider the possibility of flipping no rows or columns.
Then, consider flipping only the first row, or only the second row, and so on, trying each row individually.
Next, consider flipping every pair of rows, then every group of three rows, and so on, until you've considered flipping all possible combinations of rows.
For each of these row-flipping scenarios, repeat the same process for the columns. Try flipping no columns, then each column individually, then every pair of columns, and so on, until you've tried all combinations of columns.
After flipping a specific set of rows and columns, check if the entire grid now contains only zeros.
If the grid contains only zeros, remember this combination as a valid solution.
After exploring all possible combinations of row and column flips, determine if any valid solutions were found. If a solution was found, the grid can be converted to all zeroes.

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):
            flipped_grid = [row[:] for row in grid]

            # Flip rows based on the row_mask
            for row_index in range(number_of_rows):
                if (row_mask >> row_index) & 1:
                    for column_index in range(number_of_columns):
                        flipped_grid[row_index][column_index] = 1 - flipped_grid[row_index][column_index]

            # Flip columns based on the column_mask
            for column_index in range(number_of_columns):
                if (column_mask >> column_index) & 1:
                    for row_index in range(number_of_rows):
                        flipped_grid[row_index][column_index] = 1 - flipped_grid[row_index][column_index]

            # Check if all elements are now zero
            all_zeros = True
            for row_index in range(number_of_rows):
                for column_index in range(number_of_columns):
                    if flipped_grid[row_index][column_index] == 1:
                        all_zeros = False
                        break
                if not all_zeros:
                    break

            # If a solution is found, return True
            if all_zeros:
                return True

    # If no solution found, return False
    return False

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Check if any two rows are inherently contradictory. Two rows are contradictory if, no matter how columns are flipped, you can't make them the same.
Specifically, for each pair of rows, see if either the rows are identical or if they are the inverse of each other (where every cell in one row is the opposite of the other).
If any row pair is neither identical nor inverse, then there's no way to make the whole matrix all zeros by flipping rows and columns, so you can immediately say it's not possible.
If all pairs of rows are either identical or inverse, it means a sequence of row and column flips could make the whole grid zero, so the answer is yes.

Code Implementation

def solve():
    matrix = [
        [0, 1, 0],
        [1, 0, 1],
        [0, 1, 0]
    ]

    number_of_rows = len(matrix)

    # Check if all pairs of rows are identical or inverse.
    for i in range(number_of_rows):
        for j in range(i + 1, number_of_rows):
            rows_are_identical = True
            rows_are_inverse = True

            for column_index in range(len(matrix[0])):
                if matrix[i][column_index] != matrix[j][column_index]:
                    rows_are_identical = False

                if matrix[i][column_index] == matrix[j][column_index]:
                    rows_are_inverse = False

            # If the rows are neither identical nor inverse, return false
            if not rows_are_identical and not rows_are_inverse:
                return False

    # All rows are either identical or inverse.
    return True

def remove_all_ones(matrix):
    number_of_rows = len(matrix)
    if number_of_rows <= 1:
        return True

    #Iterate through all possible row pairs
    for row_index_1 in range(number_of_rows):
        for row_index_2 in range(row_index_1 + 1, number_of_rows):

            rows_are_identical = True
            rows_are_inverse = True
            number_of_columns = len(matrix[0])

            for column_index in range(number_of_columns):
                if matrix[row_index_1][column_index] != matrix[row_index_2][column_index]:
                    rows_are_identical = False

                if matrix[row_index_1][column_index] == matrix[row_index_2][column_index]:
                    rows_are_inverse = False
            
            #Important: If rows aren't identical/inverse, transformation is impossible
            if not rows_are_identical and not rows_are_inverse:
                return False

    return True

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty matrix	Return true if the matrix is null or empty, as no '1's exist, thus satisfying the condition.
Matrix with only one row or one column	Check if a single row or column contains only 0s after potential flipping; if so, return true; otherwise, return false.
Matrix with all zeros	Return true immediately as the condition is already met (no ones to remove).
Matrix with all ones	Test all combinations of row and column flips to determine if a valid solution exists; if not, return false.
Large matrix (scalability)	Ensure the algorithm's time complexity is optimized (e.g., avoiding exhaustive search when possible) to handle large inputs within reasonable time limits.
A case where no solution exists, i.e., no combination of flips leads to all zeros	The algorithm should correctly identify when no solution is possible and return false.
Integer overflow when calculating row/column flips	Use appropriate data types (e.g., long) to prevent integer overflow when counting the number of '1's in rows and columns.
Matrix with mostly ones and very few zeros	This input can cause the algorithm to explore many invalid combinations of row/column flips, so efficient pruning/backtracking is needed to improve performance.

Null or empty matrix

How to Handle:

Return true if the matrix is null or empty, as no '1's exist, thus satisfying the condition.

Matrix with only one row or one column

How to Handle:

Check if a single row or column contains only 0s after potential flipping; if so, return true; otherwise, return false.

Matrix with all zeros

How to Handle:

Return true immediately as the condition is already met (no ones to remove).

Matrix with all ones

How to Handle:

Test all combinations of row and column flips to determine if a valid solution exists; if not, return false.

Large matrix (scalability)

How to Handle:

Ensure the algorithm's time complexity is optimized (e.g., avoiding exhaustive search when possible) to handle large inputs within reasonable time limits.

A case where no solution exists, i.e., no combination of flips leads to all zeros

How to Handle:

The algorithm should correctly identify when no solution is possible and return false.

Integer overflow when calculating row/column flips

How to Handle:

Use appropriate data types (e.g., long) to prevent integer overflow when counting the number of '1's in rows and columns.

Matrix with mostly ones and very few zeros

How to Handle:

This input can cause the algorithm to explore many invalid combinations of row/column flips, so efficient pruning/backtracking is needed to improve performance.