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Score After Flipping Matrix

#1884 Most AskedMedium

Score After Flipping Matrix #1884 Most Asked

Medium

Google

9 views

Topics:

Greedy AlgorithmsBit Manipulation

You are given an m x n binary matrix grid.

A move consists of choosing any row or column and toggling each value in that row or column (i.e., changing all 0's to 1's, and all 1's to 0's).

Every row of the matrix is interpreted as a binary number, and the score of the matrix is the sum of these numbers.

Return the highest possible score after making any number of moves (including zero moves).

Example 1:

Input: grid = [[0,0,1,1],[1,0,1,0],[1,1,0,0]]
Output: 39
Explanation: 0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39

Example 2:

Input: grid = [[0]]
Output: 1

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 20
grid[i][j] is either 0 or 1.

Try coding it in LeetCode

Score After Flipping Matrix

#1884 Most AskedMedium

Score After Flipping Matrix #1884 Most Asked

Medium

Google

9 views

Topics:

Greedy AlgorithmsBit Manipulation

You are given an m x n binary matrix grid.

A move consists of choosing any row or column and toggling each value in that row or column (i.e., changing all 0's to 1's, and all 1's to 0's).

Every row of the matrix is interpreted as a binary number, and the score of the matrix is the sum of these numbers.

Return the highest possible score after making any number of moves (including zero moves).

Example 1:

Input: grid = [[0,0,1,1],[1,0,1,0],[1,1,0,0]]
Output: 39
Explanation: 0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39

Example 2:

Input: grid = [[0]]
Output: 1

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 20
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 (number of rows and columns)? Are there any constraints on these dimensions?
What is the range of values within the matrix? Can they be negative, zero, or floating-point numbers?
By 'flipping,' do you mean inverting the bits (0 to 1, 1 to 0) within the row or column, or swapping the positions of the elements?
If after flipping rows and columns, multiple outcomes yield the same maximum score, is any one of those outcomes acceptable?
What should I return if the input matrix is null or empty?

Brute Force Solution

Approach

The goal is to maximize the score of a grid by flipping rows and columns. The brute force method explores every possible combination of flips to find the best score.

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

Consider all possible combinations of flipping or not flipping each row. Think of it like trying all possible switches for each row.
For each of those row flip combinations, look at the resulting grid.
Then, for each resulting grid after row flips, consider all possible combinations of flipping or not flipping each column. Again, think of trying all switch options for each column.
After each row and column flip combination, calculate the score of the resulting grid.
Compare all the calculated scores from all the combinations of row and column flips.
The highest score you find across all these combinations is the answer.

Code Implementation

def score_after_flipping_matrix_brute_force(matrix):
    number_of_rows = len(matrix)
    number_of_cols = len(matrix[0])
    maximum_score = 0

    # Iterate through all possible row flip combinations.
    for row_mask in range(2**number_of_rows):
        flipped_matrix = []
        for row_index in range(number_of_rows):
            if (row_mask >> row_index) & 1:
                flipped_matrix.append([1 - bit for bit in matrix[row_index]])
            else:
                flipped_matrix.append(matrix[row_index])

        # Iterate through all possible column flip combinations.
        for column_mask in range(2**number_of_cols):
            temp_matrix = []
            for row_index in range(number_of_rows):
                new_row = []
                for col_index in range(number_of_cols):
                    if (column_mask >> col_index) & 1:
                        new_row.append(1 - flipped_matrix[row_index][col_index])
                    else:
                        new_row.append(flipped_matrix[row_index][col_index])
                temp_matrix.append(new_row)

            # Calculate the score for the current matrix.
            current_score = 0
            for row in temp_matrix:
                row_string = ''.join(map(str, row))
                current_score += int(row_string, 2)

            # Update the maximum score if necessary.
            maximum_score = max(maximum_score, current_score)

    return maximum_score

Big(O) Analysis

Time Complexity

O(2^(n+m)) – Let n be the number of rows and m be the number of columns in the grid. The algorithm considers all possible combinations of row flips, which takes O(2^n) time. For each row flip combination, it considers all possible column flip combinations, which takes O(2^m) time. Therefore, the total time complexity is O(2^n * 2^m), which can be simplified to O(2^(n+m)). The time complexity grows exponentially with the sum of the number of rows and columns due to the exhaustive search of all possible flip combinations for both rows and columns.

Space Complexity

O(1) – The brute force method described explores all possible combinations of row and column flips without storing the intermediate grids or combinations explicitly. It calculates the score for each combination on the fly and keeps track of only the maximum score seen so far. Therefore, the auxiliary space used is constant, regardless of the size of the input matrix.

Optimal Solution

Approach

The key to maximizing the score is to ensure that the most significant bit of each number in the matrix is a 1. This is done by strategically flipping rows and then columns to achieve this goal. The process prioritizes making the leftmost bit a 1, and then it optimizes the score further.

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

First, check each row. If the first number in a row is not a 1, flip the entire row. This ensures that every number has a 1 in its most significant position.
Next, go through each column, except the first one. For each column, count how many 1s are in that column and how many 0s.
If a column has more 0s than 1s, flip that entire column. This increases the overall score because it maximizes the number of 1s in that column.
Finally, calculate the score of the modified matrix. This is the sum of all the numbers in the matrix, treating each row as a binary number.

Code Implementation

def score_after_flipping_matrix(matrix):
    number_of_rows = len(matrix)
    number_of_columns = len(matrix[0])

    for row_index in range(number_of_rows):
        # Ensure the most significant bit is 1 for each row.
        if matrix[row_index][0] == 0:
            for column_index in range(number_of_columns):
                matrix[row_index][column_index] ^= 1

    for column_index in range(1, number_of_columns):
        ones_count = 0
        zeros_count = 0

        for row_index in range(number_of_rows):
            if matrix[row_index][column_index] == 1:
                ones_count += 1
            else:
                zeros_count += 1

        # Flip the column if zeros are more than ones to maximize score.
        if zeros_count > ones_count:
            for row_index in range(number_of_rows):
                matrix[row_index][column_index] ^= 1

    total_score = 0
    for row_index in range(number_of_rows):
        row_value = 0
        for column_index in range(number_of_columns):
            row_value = (row_value * 2) + matrix[row_index][column_index]
        total_score += row_value

    return total_score

Big(O) Analysis

Time Complexity

O(m * n) – The algorithm iterates through each row (m rows) to potentially flip it based on the first element, taking O(m * n) time. Then it iterates through n-1 columns, and for each column, it iterates through m rows to count the number of 0s and 1s, also taking O(m * n) time. Finally, calculating the score iterates through all m * n elements. Thus, the dominant time complexity is proportional to m * n operations which results in O(m * n).

Space Complexity

O(1) – The algorithm primarily uses in-place operations on the input matrix. It only requires a few constant space variables for counting 1s and 0s in each column. No auxiliary data structures that scale with the input matrix size are created. Therefore, the space complexity remains constant, regardless of the matrix dimensions.

Edge Cases

Case	How to Handle
Null or empty matrix input	Return 0, indicating no score possible.
Matrix with only one row or one column	If one row, flip if the first element is 0; If one column, flip each row to maximize the most significant bit.
Matrix with all 0s or all 1s	If all 0s, flip rows to make first column 1s; If all 1s, no flipping needed.
Large matrix (performance considerations)	Ensure the solution has O(m*n) time complexity where m is the number of rows and n is the number of columns to scale efficiently.
Matrix with negative numbers (if allowed based on extended problem description)	If negatives are allowed, adjust the flipping logic to maximize the sum after flipping, potentially more complex.
Integer overflow when calculating the score	Use a larger data type (e.g., long) to store the intermediate results of row sums to prevent overflow if applicable.
Maximum integer value (Integer.MAX_VALUE) in the input matrix	Handle the maximum integer values based on the score calculating formula, to prevent overflow.
Matrix dimensions are 1x1	If the single element is 0, flip to 1; otherwise do nothing.

Null or empty matrix input

How to Handle:

Return 0, indicating no score possible.

Matrix with only one row or one column

How to Handle:

If one row, flip if the first element is 0; If one column, flip each row to maximize the most significant bit.

Matrix with all 0s or all 1s

How to Handle:

If all 0s, flip rows to make first column 1s; If all 1s, no flipping needed.

Large matrix (performance considerations)

How to Handle:

Ensure the solution has O(m*n) time complexity where m is the number of rows and n is the number of columns to scale efficiently.

Matrix with negative numbers (if allowed based on extended problem description)

How to Handle:

If negatives are allowed, adjust the flipping logic to maximize the sum after flipping, potentially more complex.

Integer overflow when calculating the score

How to Handle:

Use a larger data type (e.g., long) to store the intermediate results of row sums to prevent overflow if applicable.

Maximum integer value (Integer.MAX_VALUE) in the input matrix

How to Handle:

Handle the maximum integer values based on the score calculating formula, to prevent overflow.

Matrix dimensions are 1x1

How to Handle:

If the single element is 0, flip to 1; otherwise do nothing.

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 (number of rows and columns)? Are there any constraints on these dimensions?
What is the range of values within the matrix? Can they be negative, zero, or floating-point numbers?
By 'flipping,' do you mean inverting the bits (0 to 1, 1 to 0) within the row or column, or swapping the positions of the elements?
If after flipping rows and columns, multiple outcomes yield the same maximum score, is any one of those outcomes acceptable?
What should I return if the input matrix is null or empty?

Brute Force Solution

Approach

The goal is to maximize the score of a grid by flipping rows and columns. The brute force method explores every possible combination of flips to find the best score.

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

Consider all possible combinations of flipping or not flipping each row. Think of it like trying all possible switches for each row.
For each of those row flip combinations, look at the resulting grid.
Then, for each resulting grid after row flips, consider all possible combinations of flipping or not flipping each column. Again, think of trying all switch options for each column.
After each row and column flip combination, calculate the score of the resulting grid.
Compare all the calculated scores from all the combinations of row and column flips.
The highest score you find across all these combinations is the answer.

Code Implementation

def score_after_flipping_matrix_brute_force(matrix):
    number_of_rows = len(matrix)
    number_of_cols = len(matrix[0])
    maximum_score = 0

    # Iterate through all possible row flip combinations.
    for row_mask in range(2**number_of_rows):
        flipped_matrix = []
        for row_index in range(number_of_rows):
            if (row_mask >> row_index) & 1:
                flipped_matrix.append([1 - bit for bit in matrix[row_index]])
            else:
                flipped_matrix.append(matrix[row_index])

        # Iterate through all possible column flip combinations.
        for column_mask in range(2**number_of_cols):
            temp_matrix = []
            for row_index in range(number_of_rows):
                new_row = []
                for col_index in range(number_of_cols):
                    if (column_mask >> col_index) & 1:
                        new_row.append(1 - flipped_matrix[row_index][col_index])
                    else:
                        new_row.append(flipped_matrix[row_index][col_index])
                temp_matrix.append(new_row)

            # Calculate the score for the current matrix.
            current_score = 0
            for row in temp_matrix:
                row_string = ''.join(map(str, row))
                current_score += int(row_string, 2)

            # Update the maximum score if necessary.
            maximum_score = max(maximum_score, current_score)

    return maximum_score

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, check each row. If the first number in a row is not a 1, flip the entire row. This ensures that every number has a 1 in its most significant position.
Next, go through each column, except the first one. For each column, count how many 1s are in that column and how many 0s.
If a column has more 0s than 1s, flip that entire column. This increases the overall score because it maximizes the number of 1s in that column.
Finally, calculate the score of the modified matrix. This is the sum of all the numbers in the matrix, treating each row as a binary number.

Code Implementation

def score_after_flipping_matrix(matrix):
    number_of_rows = len(matrix)
    number_of_columns = len(matrix[0])

    for row_index in range(number_of_rows):
        # Ensure the most significant bit is 1 for each row.
        if matrix[row_index][0] == 0:
            for column_index in range(number_of_columns):
                matrix[row_index][column_index] ^= 1

    for column_index in range(1, number_of_columns):
        ones_count = 0
        zeros_count = 0

        for row_index in range(number_of_rows):
            if matrix[row_index][column_index] == 1:
                ones_count += 1
            else:
                zeros_count += 1

        # Flip the column if zeros are more than ones to maximize score.
        if zeros_count > ones_count:
            for row_index in range(number_of_rows):
                matrix[row_index][column_index] ^= 1

    total_score = 0
    for row_index in range(number_of_rows):
        row_value = 0
        for column_index in range(number_of_columns):
            row_value = (row_value * 2) + matrix[row_index][column_index]
        total_score += row_value

    return total_score

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty matrix input	Return 0, indicating no score possible.
Matrix with only one row or one column	If one row, flip if the first element is 0; If one column, flip each row to maximize the most significant bit.
Matrix with all 0s or all 1s	If all 0s, flip rows to make first column 1s; If all 1s, no flipping needed.
Large matrix (performance considerations)	Ensure the solution has O(m*n) time complexity where m is the number of rows and n is the number of columns to scale efficiently.
Matrix with negative numbers (if allowed based on extended problem description)	If negatives are allowed, adjust the flipping logic to maximize the sum after flipping, potentially more complex.
Integer overflow when calculating the score	Use a larger data type (e.g., long) to store the intermediate results of row sums to prevent overflow if applicable.
Maximum integer value (Integer.MAX_VALUE) in the input matrix	Handle the maximum integer values based on the score calculating formula, to prevent overflow.
Matrix dimensions are 1x1	If the single element is 0, flip to 1; otherwise do nothing.

Null or empty matrix input

How to Handle:

Return 0, indicating no score possible.

Matrix with only one row or one column

How to Handle:

If one row, flip if the first element is 0; If one column, flip each row to maximize the most significant bit.

Matrix with all 0s or all 1s

How to Handle:

If all 0s, flip rows to make first column 1s; If all 1s, no flipping needed.

Large matrix (performance considerations)

How to Handle:

Ensure the solution has O(m*n) time complexity where m is the number of rows and n is the number of columns to scale efficiently.

Matrix with negative numbers (if allowed based on extended problem description)

How to Handle:

If negatives are allowed, adjust the flipping logic to maximize the sum after flipping, potentially more complex.

Integer overflow when calculating the score

How to Handle:

Use a larger data type (e.g., long) to store the intermediate results of row sums to prevent overflow if applicable.

Maximum integer value (Integer.MAX_VALUE) in the input matrix

How to Handle:

Handle the maximum integer values based on the score calculating formula, to prevent overflow.

Matrix dimensions are 1x1

How to Handle:

If the single element is 0, flip to 1; otherwise do nothing.

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