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Flip Columns For Maximum Number of Equal Rows

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ArraysBit Manipulation

You are given an m x n binary matrix matrix.

You can choose any number of columns in the matrix and flip every cell in that column (i.e., Change the value of the cell from 0 to 1 or vice versa).

Return the maximum number of rows that have all values equal after some number of flips.

Example 1:

Input: matrix = [[0,1],[1,1]]
Output: 1
Explanation: After flipping no values, 1 row has all values equal.

Example 2:

Input: matrix = [[0,1],[1,0]]
Output: 2
Explanation: After flipping values in the first column, both rows have equal values.

Example 3:

Input: matrix = [[0,0,0],[0,0,1],[1,1,0]]
Output: 2
Explanation: After flipping values in the first two columns, the last two rows have equal values.

Constraints:

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

Try coding it in LeetCode

Flip Columns For Maximum Number of Equal Rows

Medium

Asked by:

25 views

Topics:

ArraysBit Manipulation

You are given an m x n binary matrix matrix.

You can choose any number of columns in the matrix and flip every cell in that column (i.e., Change the value of the cell from 0 to 1 or vice versa).

Return the maximum number of rows that have all values equal after some number of flips.

Example 1:

Input: matrix = [[0,1],[1,1]]
Output: 1
Explanation: After flipping no values, 1 row has all values equal.

Example 2:

Input: matrix = [[0,1],[1,0]]
Output: 2
Explanation: After flipping values in the first column, both rows have equal values.

Example 3:

Input: matrix = [[0,0,0],[0,0,1],[1,1,0]]
Output: 2
Explanation: After flipping values in the first two columns, the last two rows have equal values.

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 300
matrix[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 and value ranges of the 2D array? Specifically, what are the maximum values for the number of rows and columns, and what's the range for the elements within each row (e.g., are they only 0 and 1)?
If no number of column flips results in at least one identical row, what value should I return? Should I return 0?
Can a row contain all zeros or all ones initially? And how should I handle a completely empty input array (zero rows or zero columns)?
Are all rows guaranteed to have the same number of columns, or could there be rows of varying lengths? If so, what should I do?
Does the order of rows matter? In other words, are we only concerned about the number of identical rows after flips, or must they also be in a particular sequence?

Brute Force Solution

Approach

The brute force method to maximize equal rows after column flips involves trying every possible combination of flips. We systematically go through each column and consider both flipping it and not flipping it, checking the resulting matrix to see how many rows are equal. The best outcome across all flip combinations is then chosen.

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

Start by considering the first column.
We have two choices: either flip it or don't flip it.
If we flip it, change all the values in that column (0 becomes 1 and 1 becomes 0).
Now look at the second column. Again, we can flip it or leave it alone. This doubles the number of possibilities we need to check.
Keep going through each column, one at a time. Each time, explore the two choices: flip the column or leave it as is.
After making a choice for every column, we have one possible arrangement of flipped and non-flipped columns.
For this arrangement, go through each row and compare it to every other row to see how many rows are identical.
Keep track of the arrangement that results in the highest number of equal rows.
Repeat this process for every possible arrangement of column flips.
Finally, choose the arrangement of column flips that gave us the most equal rows. That's our answer.

Code Implementation

def flip_columns_for_maximum_equal_rows_brute_force(matrix):
    number_of_rows = len(matrix)
    number_of_columns = len(matrix[0]) if number_of_rows > 0 else 0

    maximum_equal_rows = 0

    # Iterate through all possible column flip combinations.
    for i in range(2 ** number_of_columns):
        flipped_matrix = []
        for row in matrix:
            flipped_matrix.append(row[:])

        # Determine which columns to flip based on the current combination.
        for column_index in range(number_of_columns):
            if (i >> column_index) & 1:
                # Flip the column.

                for row_index in range(number_of_rows):
                    flipped_matrix[row_index][column_index] = 1 - flipped_matrix[row_index][column_index]

        equal_rows_count = 0
        # Count the number of equal rows in the flipped matrix.
        for row_index in range(number_of_rows):
            current_row = flipped_matrix[row_index]
            current_equal_rows = 1
            for other_row_index in range(row_index + 1, number_of_rows):
                if flipped_matrix[other_row_index] == current_row:
                    current_equal_rows += 1

            equal_rows_count = equal_rows_count + current_equal_rows

        #Update maximum rows
        maximum_equal_rows = max(maximum_equal_rows, equal_rows_count)

    return maximum_equal_rows

Big(O) Analysis

Time Complexity

O(2^C * R^2) – The algorithm explores every possible combination of column flips. Since there are C columns, this generates 2^C possibilities. For each of these 2^C configurations, it iterates through all rows (R) and for each row, it compares it with every other row which involves another loop through R rows to check for equality. Therefore, for each configuration, we have an R*R operation. Multiplying the number of configurations by the row comparison complexity, we get 2^C * R * R, which simplifies to O(2^C * R^2).

Space Complexity

O(1) – The brute force approach, as described, doesn't utilize any auxiliary data structures that scale with the size of the input matrix (number of rows or columns). It operates directly on the input matrix and uses a constant amount of extra space for bookkeeping, such as storing the current maximum number of equal rows found so far. Therefore, the space complexity remains constant irrespective of the input size.

Optimal Solution

Approach

The core idea is to realize that rows can only be equal if they are either exactly the same, or the exact opposite of each other. Instead of brute-forcing all column flips, we'll focus on identifying these row pairs and count how many equal rows we can achieve.

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

First, pick a row to be your 'reference' row. It doesn't matter which one.
Then, for every other row in your group, check if it's either exactly the same as your reference row or exactly the opposite (where every value is flipped).
If a row is exactly the same or the exact opposite of the reference row, then you can make it equal by either doing nothing (if it's the same) or flipping all the columns that need to be flipped to make it the same.
Keep track of how many rows you can make equal to your reference row.
Repeat this process, using each row as the reference row once.
The biggest number of equal rows you can make out of all the times you used a reference row is your answer.

Code Implementation

def flip_columns_for_maximum_equal_rows(matrix):
    maximum_equal_rows = 0
    number_of_rows = len(matrix)

    for row_index in range(number_of_rows):
        reference_row = matrix[row_index]
        equal_row_count = 0

        # Use each row as a reference row.
        for other_row_index in range(number_of_rows):
            other_row = matrix[other_row_index]
            is_same = True
            is_opposite = True

            # Check if rows are the same or opposite.
            for col_index in range(len(reference_row)):
                if reference_row[col_index] != other_row[col_index]:
                    is_same = False
                if reference_row[col_index] == other_row[col_index]:
                    is_opposite = False

            if is_same or is_opposite:
                # Increment count if rows are the same or opposite.
                equal_row_count += 1

        # Keep track of the maximum equal rows.
        maximum_equal_rows = max(maximum_equal_rows, equal_row_count)

    return maximum_equal_rows

Big(O) Analysis

Time Complexity

O(m*n) – The algorithm iterates through each row (m rows) and compares it with every other row (m rows). For each pair of rows, it compares the values of each column (n columns) to determine if they are the same or opposites. Thus, the comparison operation inside the nested loop takes O(n) time. Since we have nested loops iterating through m rows each, and each comparison takes O(n) time, the total time complexity is O(m*m*n). Because we must iterate through each row and column, the algorithm will always take a minimum of O(m*n). In a simplified scenario, without the necessity of comparison, the time complexity might be O(m*n). However, the comparisons are essential and occur within the loop, and the overall operation approximates to O(m*n) as it is the dominant factor.

Space Complexity

O(1) – The algorithm iterates through the rows of the input matrix, comparing each row to a chosen reference row. It doesn't create any auxiliary data structures whose size depends on the number of rows or columns of the input. The comparison involves checking if a row is the same or the opposite of the reference row, which can be done in place or with a few constant-sized variables. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Empty matrix (rows or columns)	Return 0, as there are no rows to make equal.
Matrix with only one row	Return 1, as the single row is already equal to itself.
Matrix with only one column	Return the number of rows, as all rows will be equal after potentially flipping the single column.
All rows are identical	Return the number of rows, as no flipping is required.
All rows are bitwise complements of each other	Return the number of rows, as flipping any column will make all rows equal.
Matrix with a large number of rows and columns (scalability)	Using a hash map to store row patterns avoids pairwise comparisons and allows efficient scaling for large inputs.
Rows that become identical after flipping different columns	The hash map correctly identifies these rows based on their flipped pattern.
Integer overflow when calculating hash for row patterns	Choose a hash function that minimizes collisions and avoids integer overflow issues.

Empty matrix (rows or columns)

How to Handle:

Return 0, as there are no rows to make equal.

Matrix with only one row

How to Handle:

Return 1, as the single row is already equal to itself.

Matrix with only one column

How to Handle:

Return the number of rows, as all rows will be equal after potentially flipping the single column.

All rows are identical

How to Handle:

Return the number of rows, as no flipping is required.

All rows are bitwise complements of each other

How to Handle:

Return the number of rows, as flipping any column will make all rows equal.

Matrix with a large number of rows and columns (scalability)

How to Handle:

Using a hash map to store row patterns avoids pairwise comparisons and allows efficient scaling for large inputs.

Rows that become identical after flipping different columns

How to Handle:

The hash map correctly identifies these rows based on their flipped pattern.

Integer overflow when calculating hash for row patterns

How to Handle:

Choose a hash function that minimizes collisions and avoids integer overflow issues.

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 and value ranges of the 2D array? Specifically, what are the maximum values for the number of rows and columns, and what's the range for the elements within each row (e.g., are they only 0 and 1)?
If no number of column flips results in at least one identical row, what value should I return? Should I return 0?
Can a row contain all zeros or all ones initially? And how should I handle a completely empty input array (zero rows or zero columns)?
Are all rows guaranteed to have the same number of columns, or could there be rows of varying lengths? If so, what should I do?
Does the order of rows matter? In other words, are we only concerned about the number of identical rows after flips, or must they also be in a particular sequence?

Brute Force Solution

Approach

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

Start by considering the first column.
We have two choices: either flip it or don't flip it.
If we flip it, change all the values in that column (0 becomes 1 and 1 becomes 0).
Now look at the second column. Again, we can flip it or leave it alone. This doubles the number of possibilities we need to check.
Keep going through each column, one at a time. Each time, explore the two choices: flip the column or leave it as is.
After making a choice for every column, we have one possible arrangement of flipped and non-flipped columns.
For this arrangement, go through each row and compare it to every other row to see how many rows are identical.
Keep track of the arrangement that results in the highest number of equal rows.
Repeat this process for every possible arrangement of column flips.
Finally, choose the arrangement of column flips that gave us the most equal rows. That's our answer.

Code Implementation

def flip_columns_for_maximum_equal_rows_brute_force(matrix):
    number_of_rows = len(matrix)
    number_of_columns = len(matrix[0]) if number_of_rows > 0 else 0

    maximum_equal_rows = 0

    # Iterate through all possible column flip combinations.
    for i in range(2 ** number_of_columns):
        flipped_matrix = []
        for row in matrix:
            flipped_matrix.append(row[:])

        # Determine which columns to flip based on the current combination.
        for column_index in range(number_of_columns):
            if (i >> column_index) & 1:
                # Flip the column.

                for row_index in range(number_of_rows):
                    flipped_matrix[row_index][column_index] = 1 - flipped_matrix[row_index][column_index]

        equal_rows_count = 0
        # Count the number of equal rows in the flipped matrix.
        for row_index in range(number_of_rows):
            current_row = flipped_matrix[row_index]
            current_equal_rows = 1
            for other_row_index in range(row_index + 1, number_of_rows):
                if flipped_matrix[other_row_index] == current_row:
                    current_equal_rows += 1

            equal_rows_count = equal_rows_count + current_equal_rows

        #Update maximum rows
        maximum_equal_rows = max(maximum_equal_rows, equal_rows_count)

    return maximum_equal_rows

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, pick a row to be your 'reference' row. It doesn't matter which one.
Then, for every other row in your group, check if it's either exactly the same as your reference row or exactly the opposite (where every value is flipped).
If a row is exactly the same or the exact opposite of the reference row, then you can make it equal by either doing nothing (if it's the same) or flipping all the columns that need to be flipped to make it the same.
Keep track of how many rows you can make equal to your reference row.
Repeat this process, using each row as the reference row once.
The biggest number of equal rows you can make out of all the times you used a reference row is your answer.

Code Implementation

def flip_columns_for_maximum_equal_rows(matrix):
    maximum_equal_rows = 0
    number_of_rows = len(matrix)

    for row_index in range(number_of_rows):
        reference_row = matrix[row_index]
        equal_row_count = 0

        # Use each row as a reference row.
        for other_row_index in range(number_of_rows):
            other_row = matrix[other_row_index]
            is_same = True
            is_opposite = True

            # Check if rows are the same or opposite.
            for col_index in range(len(reference_row)):
                if reference_row[col_index] != other_row[col_index]:
                    is_same = False
                if reference_row[col_index] == other_row[col_index]:
                    is_opposite = False

            if is_same or is_opposite:
                # Increment count if rows are the same or opposite.
                equal_row_count += 1

        # Keep track of the maximum equal rows.
        maximum_equal_rows = max(maximum_equal_rows, equal_row_count)

    return maximum_equal_rows

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty matrix (rows or columns)	Return 0, as there are no rows to make equal.
Matrix with only one row	Return 1, as the single row is already equal to itself.
Matrix with only one column	Return the number of rows, as all rows will be equal after potentially flipping the single column.
All rows are identical	Return the number of rows, as no flipping is required.
All rows are bitwise complements of each other	Return the number of rows, as flipping any column will make all rows equal.
Matrix with a large number of rows and columns (scalability)	Using a hash map to store row patterns avoids pairwise comparisons and allows efficient scaling for large inputs.
Rows that become identical after flipping different columns	The hash map correctly identifies these rows based on their flipped pattern.
Integer overflow when calculating hash for row patterns	Choose a hash function that minimizes collisions and avoids integer overflow issues.

Empty matrix (rows or columns)

How to Handle:

Return 0, as there are no rows to make equal.

Matrix with only one row

How to Handle:

Return 1, as the single row is already equal to itself.

Matrix with only one column

How to Handle:

Return the number of rows, as all rows will be equal after potentially flipping the single column.

All rows are identical

How to Handle:

Return the number of rows, as no flipping is required.

All rows are bitwise complements of each other

How to Handle:

Return the number of rows, as flipping any column will make all rows equal.

Matrix with a large number of rows and columns (scalability)

How to Handle:

Using a hash map to store row patterns avoids pairwise comparisons and allows efficient scaling for large inputs.

Rows that become identical after flipping different columns

How to Handle:

The hash map correctly identifies these rows based on their flipped pattern.

Integer overflow when calculating hash for row patterns

How to Handle:

Choose a hash function that minimizes collisions and avoids integer overflow issues.