Jobs

Interviews

Discussions

Maximum Matrix Sum

Medium

Asked by:

22 views

Topics:

ArraysGreedy Algorithms

You are given an n x n integer matrix. You can do the following operation any number of times:

Choose any two adjacent elements of matrix and multiply each of them by -1.

Two elements are considered adjacent if and only if they share a border.

Your goal is to maximize the summation of the matrix's elements. Return the maximum sum of the matrix's elements using the operation mentioned above.

Example 1:

Input: matrix = [[1,-1],[-1,1]]
Output: 4
Explanation: We can follow the following steps to reach sum equals 4:
- Multiply the 2 elements in the first row by -1.
- Multiply the 2 elements in the first column by -1.

Example 2:

Input: matrix = [[1,2,3],[-1,-2,-3],[1,2,3]]
Output: 16
Explanation: We can follow the following step to reach sum equals 16:
- Multiply the 2 last elements in the second row by -1.

Constraints:

n == matrix.length == matrix[i].length
2 <= n <= 250
-10⁵ <= matrix[i][j] <= 10⁵

Try coding it in LeetCode

Maximum Matrix Sum

Medium

Asked by:

22 views

Topics:

ArraysGreedy Algorithms

You are given an n x n integer matrix. You can do the following operation any number of times:

Choose any two adjacent elements of matrix and multiply each of them by -1.

Two elements are considered adjacent if and only if they share a border.

Your goal is to maximize the summation of the matrix's elements. Return the maximum sum of the matrix's elements using the operation mentioned above.

Example 1:

Input: matrix = [[1,-1],[-1,1]]
Output: 4
Explanation: We can follow the following steps to reach sum equals 4:
- Multiply the 2 elements in the first row by -1.
- Multiply the 2 elements in the first column by -1.

Example 2:

Input: matrix = [[1,2,3],[-1,-2,-3],[1,2,3]]
Output: 16
Explanation: We can follow the following step to reach sum equals 16:
- Multiply the 2 last elements in the second row by -1.

Constraints:

n == matrix.length == matrix[i].length
2 <= n <= 250
-10⁵ <= matrix[i][j] <= 10⁵

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 is the range of values that each element in the matrix can have? Can elements be negative, zero, or only positive?
Can the input matrix be empty or null? What should I return in these cases?
By 'maximum sum', do you mean the absolute maximum sum achievable by converting negative numbers to positive numbers according to the rules, or is there some other interpretation?
If there are multiple ways to achieve the maximum sum, does it matter which combination of flips I use, or can I return any valid result?
Can you provide a small example input and the expected output to illustrate the problem more clearly?

Brute Force Solution

Approach

To find the largest sum of a matrix by brute force, we essentially check every single possible combination of signs for each number in the matrix. We aim to maximize the sum by strategically flipping signs to eliminate negative numbers.

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

Consider that each number in the matrix can either be positive or negative.
Imagine going through every single possibility of positive or negative signs for each number in the matrix.
For each of these possibilities, calculate the sum of all the numbers in the matrix.
Keep track of the largest sum that you find from all of these calculations.
After checking all possible combinations of positive and negative signs, the largest sum you've kept track of is your answer.

Code Implementation

def maximum_matrix_sum_brute_force(matrix):
    number_of_rows = len(matrix)
    number_of_columns = len(matrix[0])
    total_elements = number_of_rows * number_of_columns
    maximum_sum = float('-inf')

    # Iterate through all possible sign combinations.
    for i in range(2 ** total_elements):
        current_sum = 0
        temp_matrix = []

        bitmask = bin(i)[2:].zfill(total_elements)
        element_index = 0
        
        for row in range(number_of_rows):
            temp_row = []
            for column in range(number_of_columns):
                # Determine the sign based on the bitmask.
                if bitmask[element_index] == '0':
                    temp_row.append(matrix[row][column])
                    current_sum += matrix[row][column]
                else:
                    temp_row.append(-matrix[row][column])
                    current_sum += -matrix[row][column]

                element_index += 1
            temp_matrix.append(temp_row)

        # Update maximum sum if current sum is larger.
        if current_sum > maximum_sum:
            maximum_sum = current_sum

    return maximum_sum

Big(O) Analysis

Time Complexity

O(2^(n*m)) – Given an n x m matrix, the brute force approach involves considering all possible sign combinations for each element. Each element can be either positive or negative, leading to 2 possibilities per element. Since there are n * m elements in the matrix, the total number of combinations to check is 2^(n*m). Therefore, the time complexity is O(2^(n*m)), as we iterate through all these combinations to find the maximum sum.

Space Complexity

O(1) – The plain English description outlines a brute-force approach that imagines trying all possible sign combinations without explicitly storing each combination. It calculates the sum for each combination and tracks only the maximum sum found so far. Therefore, the only auxiliary space used is for the 'largest sum' variable, which requires constant space regardless of the input matrix size N (where N represents the total number of elements in the matrix), and possibly a temporary variable to hold the current sum. This constant space usage results in a space complexity of O(1).

Optimal Solution

Approach

The goal is to maximize the sum of the matrix by cleverly manipulating negative numbers. The trick is to realize that pairs of negative numbers can be converted to positive numbers by flipping their signs strategically, and the number of negative numbers are important to the calculation.

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

First, count how many negative numbers are in the matrix.
Calculate the total sum of the absolute values of all the numbers in the matrix. This is like assuming all numbers are positive.
If there are an even number of negative numbers, you can make them all positive with a series of sign flips. The maximum sum is simply the total absolute value sum calculated earlier.
If there are an odd number of negative numbers, you can't get rid of all the negative signs. You'll be left with one negative number.
To minimize the impact of the leftover negative number, find the smallest absolute value in the entire matrix.
Subtract twice the smallest absolute value from the total absolute value sum. This is because you are essentially changing the sign of the smallest number from positive to negative, reducing the total by twice the number's value.

Code Implementation

def maximum_matrix_sum(matrix):
    negative_number_count = 0
    absolute_sum = 0
    minimum_absolute_value = float('inf')

    for row in matrix:
        for number in row:
            if number < 0:
                negative_number_count += 1

            absolute_value = abs(number)
            absolute_sum += absolute_value
            minimum_absolute_value = min(minimum_absolute_value, absolute_value)

    # Even number of negatives can be flipped to positive.
    if negative_number_count % 2 == 0:
        return absolute_sum

    # Odd number of negatives, one will remain negative.
    # Minimize its impact by choosing smallest absolute value to negate.

    # Subtract twice the minimum absolute value to account for sign change.
    return absolute_sum - 2 * minimum_absolute_value

Big(O) Analysis

Time Complexity

O(n*m) – The algorithm iterates through all n*m elements of the matrix to count the number of negative numbers and to calculate the sum of absolute values. Then, it iterates through all n*m elements again to find the minimum absolute value. These are the only operations dependent on the input matrix. Thus, the time complexity is dominated by iterating through the matrix twice, each taking O(n*m) time. Therefore the overall time complexity is O(n*m).

Space Complexity

O(1) – The algorithm uses a constant amount of extra space. It only stores a few integer variables: one to count the number of negative numbers, one to accumulate the absolute sum, and one to keep track of the minimum absolute value. The space used by these variables does not depend on the size of the input matrix; therefore, the space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty matrix input	Return 0 immediately as there are no elements to sum.
Matrix with a single element	Return the absolute value of the single element.
Matrix with all positive numbers	Sum all elements directly as there is no benefit to flipping.
Matrix with all negative numbers	Flip all rows and columns to make all numbers positive, except for one if the matrix size is odd.
Matrix with a mix of positive, negative, and zero values	Count negative numbers, if even, sum absolute values; if odd, flip the smallest absolute value to negative.
Large matrix with integers near Integer.MAX_VALUE or Integer.MIN_VALUE, causing overflow during summation	Use long data type to store the intermediate and final sum to prevent integer overflow.
Matrix with many zeros	Zeros do not affect the sum and can be ignored when counting negative numbers and finding the smallest absolute value.
Matrix with dimensions close to maximum allowed array sizes	Ensure sufficient memory allocation and test with large inputs to prevent out-of-memory errors.

Null or empty matrix input

How to Handle:

Return 0 immediately as there are no elements to sum.

Matrix with a single element

How to Handle:

Return the absolute value of the single element.

Matrix with all positive numbers

How to Handle:

Sum all elements directly as there is no benefit to flipping.

Matrix with all negative numbers

How to Handle:

Flip all rows and columns to make all numbers positive, except for one if the matrix size is odd.

Matrix with a mix of positive, negative, and zero values

How to Handle:

Count negative numbers, if even, sum absolute values; if odd, flip the smallest absolute value to negative.

Large matrix with integers near Integer.MAX_VALUE or Integer.MIN_VALUE, causing overflow during summation

How to Handle:

Use long data type to store the intermediate and final sum to prevent integer overflow.

Matrix with many zeros

How to Handle:

Zeros do not affect the sum and can be ignored when counting negative numbers and finding the smallest absolute value.

Matrix with dimensions close to maximum allowed array sizes

How to Handle:

Ensure sufficient memory allocation and test with large inputs to prevent out-of-memory errors.

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 is the range of values that each element in the matrix can have? Can elements be negative, zero, or only positive?
Can the input matrix be empty or null? What should I return in these cases?
By 'maximum sum', do you mean the absolute maximum sum achievable by converting negative numbers to positive numbers according to the rules, or is there some other interpretation?
If there are multiple ways to achieve the maximum sum, does it matter which combination of flips I use, or can I return any valid result?
Can you provide a small example input and the expected output to illustrate the problem more clearly?

Brute Force Solution

Approach

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

Consider that each number in the matrix can either be positive or negative.
Imagine going through every single possibility of positive or negative signs for each number in the matrix.
For each of these possibilities, calculate the sum of all the numbers in the matrix.
Keep track of the largest sum that you find from all of these calculations.
After checking all possible combinations of positive and negative signs, the largest sum you've kept track of is your answer.

Code Implementation

def maximum_matrix_sum_brute_force(matrix):
    number_of_rows = len(matrix)
    number_of_columns = len(matrix[0])
    total_elements = number_of_rows * number_of_columns
    maximum_sum = float('-inf')

    # Iterate through all possible sign combinations.
    for i in range(2 ** total_elements):
        current_sum = 0
        temp_matrix = []

        bitmask = bin(i)[2:].zfill(total_elements)
        element_index = 0
        
        for row in range(number_of_rows):
            temp_row = []
            for column in range(number_of_columns):
                # Determine the sign based on the bitmask.
                if bitmask[element_index] == '0':
                    temp_row.append(matrix[row][column])
                    current_sum += matrix[row][column]
                else:
                    temp_row.append(-matrix[row][column])
                    current_sum += -matrix[row][column]

                element_index += 1
            temp_matrix.append(temp_row)

        # Update maximum sum if current sum is larger.
        if current_sum > maximum_sum:
            maximum_sum = current_sum

    return maximum_sum

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, count how many negative numbers are in the matrix.
Calculate the total sum of the absolute values of all the numbers in the matrix. This is like assuming all numbers are positive.
If there are an even number of negative numbers, you can make them all positive with a series of sign flips. The maximum sum is simply the total absolute value sum calculated earlier.
If there are an odd number of negative numbers, you can't get rid of all the negative signs. You'll be left with one negative number.
To minimize the impact of the leftover negative number, find the smallest absolute value in the entire matrix.
Subtract twice the smallest absolute value from the total absolute value sum. This is because you are essentially changing the sign of the smallest number from positive to negative, reducing the total by twice the number's value.

Code Implementation

def maximum_matrix_sum(matrix):
    negative_number_count = 0
    absolute_sum = 0
    minimum_absolute_value = float('inf')

    for row in matrix:
        for number in row:
            if number < 0:
                negative_number_count += 1

            absolute_value = abs(number)
            absolute_sum += absolute_value
            minimum_absolute_value = min(minimum_absolute_value, absolute_value)

    # Even number of negatives can be flipped to positive.
    if negative_number_count % 2 == 0:
        return absolute_sum

    # Odd number of negatives, one will remain negative.
    # Minimize its impact by choosing smallest absolute value to negate.

    # Subtract twice the minimum absolute value to account for sign change.
    return absolute_sum - 2 * minimum_absolute_value

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty matrix input	Return 0 immediately as there are no elements to sum.
Matrix with a single element	Return the absolute value of the single element.
Matrix with all positive numbers	Sum all elements directly as there is no benefit to flipping.
Matrix with all negative numbers	Flip all rows and columns to make all numbers positive, except for one if the matrix size is odd.
Matrix with a mix of positive, negative, and zero values	Count negative numbers, if even, sum absolute values; if odd, flip the smallest absolute value to negative.
Large matrix with integers near Integer.MAX_VALUE or Integer.MIN_VALUE, causing overflow during summation	Use long data type to store the intermediate and final sum to prevent integer overflow.
Matrix with many zeros	Zeros do not affect the sum and can be ignored when counting negative numbers and finding the smallest absolute value.
Matrix with dimensions close to maximum allowed array sizes	Ensure sufficient memory allocation and test with large inputs to prevent out-of-memory errors.

Null or empty matrix input

How to Handle:

Return 0 immediately as there are no elements to sum.

Matrix with a single element

How to Handle:

Return the absolute value of the single element.

Matrix with all positive numbers

How to Handle:

Sum all elements directly as there is no benefit to flipping.

Matrix with all negative numbers

How to Handle:

Flip all rows and columns to make all numbers positive, except for one if the matrix size is odd.

Matrix with a mix of positive, negative, and zero values

How to Handle:

Count negative numbers, if even, sum absolute values; if odd, flip the smallest absolute value to negative.

Large matrix with integers near Integer.MAX_VALUE or Integer.MIN_VALUE, causing overflow during summation

How to Handle:

Use long data type to store the intermediate and final sum to prevent integer overflow.

Matrix with many zeros

How to Handle:

Zeros do not affect the sum and can be ignored when counting negative numbers and finding the smallest absolute value.

Matrix with dimensions close to maximum allowed array sizes

How to Handle:

Ensure sufficient memory allocation and test with large inputs to prevent out-of-memory errors.