Difference Between Ones and Zeros in Row and Column

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

A 0-indexed m x n difference matrix diff is created with the following procedure:

Let the number of ones in the i^th row be onesRow_i.
Let the number of ones in the j^th column be onesCol_j.
Let the number of zeros in the i^th row be zerosRow_i.
Let the number of zeros in the j^th column be zerosCol_j.
diff[i][j] = onesRow_i + onesCol_j - zerosRow_i - zerosCol_j

Return the difference matrix diff.

Example 1:

Input: grid = [[0,1,1],[1,0,1],[0,0,1]]
Output: [[0,0,4],[0,0,4],[-2,-2,2]]
Explanation:
- diff[0][0] = onesRow₀ + onesCol₀ - zerosRow₀ - zerosCol₀ = 2 + 1 - 1 - 2 = 0 
- diff[0][1] = onesRow₀ + onesCol₁ - zerosRow₀ - zerosCol₁ = 2 + 1 - 1 - 2 = 0 
- diff[0][2] = onesRow₀ + onesCol₂ - zerosRow₀ - zerosCol₂ = 2 + 3 - 1 - 0 = 4 
- diff[1][0] = onesRow₁ + onesCol₀ - zerosRow₁ - zerosCol₀ = 2 + 1 - 1 - 2 = 0 
- diff[1][1] = onesRow₁ + onesCol₁ - zerosRow₁ - zerosCol₁ = 2 + 1 - 1 - 2 = 0 
- diff[1][2] = onesRow₁ + onesCol₂ - zerosRow₁ - zerosCol₂ = 2 + 3 - 1 - 0 = 4 
- diff[2][0] = onesRow₂ + onesCol₀ - zerosRow₂ - zerosCol₀ = 1 + 1 - 2 - 2 = -2
- diff[2][1] = onesRow₂ + onesCol₁ - zerosRow₂ - zerosCol₁ = 1 + 1 - 2 - 2 = -2
- diff[2][2] = onesRow₂ + onesCol₂ - zerosRow₂ - zerosCol₂ = 1 + 3 - 2 - 0 = 2

Example 2:

Input: grid = [[1,1,1],[1,1,1]]
Output: [[5,5,5],[5,5,5]]
Explanation:
- diff[0][0] = onesRow₀ + onesCol₀ - zerosRow₀ - zerosCol₀ = 3 + 2 - 0 - 0 = 5
- diff[0][1] = onesRow₀ + onesCol₁ - zerosRow₀ - zerosCol₁ = 3 + 2 - 0 - 0 = 5
- diff[0][2] = onesRow₀ + onesCol₂ - zerosRow₀ - zerosCol₂ = 3 + 2 - 0 - 0 = 5
- diff[1][0] = onesRow₁ + onesCol₀ - zerosRow₁ - zerosCol₀ = 3 + 2 - 0 - 0 = 5
- diff[1][1] = onesRow₁ + onesCol₁ - zerosRow₁ - zerosCol₁ = 3 + 2 - 0 - 0 = 5
- diff[1][2] = onesRow₁ + onesCol₂ - zerosRow₁ - zerosCol₂ = 3 + 2 - 0 - 0 = 5

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 10⁵
1 <= m * n <= 10⁵
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 is the expected range for the dimensions of the matrix (number of rows and columns)?
Will the matrix always be rectangular, or can the number of columns vary for each row?
Does the matrix only contain 0s and 1s, or are other values possible?
What is the desired output format? Should it be a new matrix of the same dimensions with the difference values, or something else?
Should I consider memory usage if the matrix is extremely large?

Brute Force Solution

Approach

The brute force way to solve this problem is to calculate the difference between the number of ones and zeros for each cell individually. This involves examining every row and every column that the cell belongs to, counting the ones and zeros.

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

For each cell in the grid, we will do the following steps.
First, we need to figure out what row the cell is in and count how many ones and zeros there are in that row.
Next, we need to figure out what column the cell is in and count how many ones and zeros there are in that column.
Now that we have the counts for both the row and the column, we can calculate the difference as described in the problem.
We can now store that difference for the current cell we are examining.
After we have done this for every cell in the grid, we will have our answer.

Code Implementation

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

    difference_grid = [[0] * number_of_columns for _ in range(number_of_rows)]

    for row in range(number_of_rows):
        for column in range(number_of_columns):
            row_ones_count = 0
            row_zeros_count = 0
            column_ones_count = 0
            column_zeros_count = 0

            # Count ones and zeros in the current row
            for column_index in range(number_of_columns):
                if grid[row][column_index] == 1:
                    row_ones_count += 1
                else:
                    row_zeros_count += 1

            # Count ones and zeros in the current column
            for row_index in range(number_of_rows):
                if grid[row_index][column] == 1:
                    column_ones_count += 1
                else:
                    column_zeros_count += 1

            # Calculate the difference and store it in the result grid
            difference_grid[row][column] = row_ones_count + column_ones_count - row_zeros_count - column_zeros_count

    return difference_grid

Big(O) Analysis

Time Complexity

O(m * n * (m + n)) – The algorithm iterates through each cell in the grid, which has dimensions m x n, resulting in m * n iterations. For each cell, the algorithm iterates through its corresponding row of length n and its corresponding column of length m to count the ones and zeros. Thus, for each cell, the row and column counts take O(m + n) time. Multiplying the iterations and cost per iteration together results in a final time complexity of O(m * n * (m + n)).

Space Complexity

O(N*M) – The brute force approach iterates through each cell of the input grid. For each cell, the algorithm calculates the difference between the number of ones and zeros in its corresponding row and column. Although not explicitly mentioned, it must store these differences for each cell in a new grid of the same size as the input grid to hold the results. Therefore, if the input grid has N rows and M columns, the algorithm requires O(N*M) space to store the difference values.

Optimal Solution

Approach

The most efficient way to solve this is to avoid checking every cell individually. Instead, we'll count the number of ones and zeros in each row and column separately. Then we'll use those counts to quickly determine the difference for each cell.

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

First, count the number of ones in each row and store these counts.
Next, count the number of ones in each column and store these counts.
Now, for each cell, we know the number of ones in its row and column.
We also know the number of zeros must be the total size of the row/column minus the number of ones.
Using these counts, we calculate the difference between the number of ones and zeros in the row and column for each cell.
Store these differences in a new grid that represents the answer.

Code Implementation

def difference_between_ones_and_zeros(grid):
    number_of_rows = len(grid)
    number_of_columns = len(grid[0])
    
    ones_in_row = [0] * number_of_rows
    ones_in_column = [0] * number_of_columns

    # Count ones in each row
    for row_index in range(number_of_rows):
        for column_index in range(number_of_columns):
            if grid[row_index][column_index] == 1:
                ones_in_row[row_index] += 1

    # Count ones in each column
    for column_index in range(number_of_columns):
        for row_index in range(number_of_rows):
            if grid[row_index][column_index] == 1:
                ones_in_column[column_index] += 1

    difference_grid = [[0] * number_of_columns for _ in range(number_of_rows)]

    # Calculate difference for each cell
    for row_index in range(number_of_rows):
        for column_index in range(number_of_columns):
            # Calculate ones and zeros based on rows and columns
            ones_row_value = ones_in_row[row_index]
            zeros_row_value = number_of_columns - ones_row_value
            ones_column_value = ones_in_column[column_index]
            zeros_column_value = number_of_rows - ones_column_value

            # Populate the difference grid
            difference_grid[row_index][column_index] = ones_row_value + ones_column_value - zeros_row_value - zeros_column_value

    return difference_grid

Big(O) Analysis

Time Complexity

O(m * n) – First, we iterate through each row to count the number of ones, which takes O(m * n) time where m is the number of rows and n is the number of columns. Next, we iterate through each column to count the number of ones, taking another O(m * n) time. Finally, we iterate through each cell again to calculate the difference based on row and column counts, adding another O(m * n) operation. Since these operations are performed sequentially, the overall time complexity is O(m * n) + O(m * n) + O(m * n), which simplifies to O(m * n).

Space Complexity

O(N) – The solution involves creating two auxiliary arrays to store the counts of ones in each row and each column. If the input grid is of size R x C, where N is the total number of cells (R * C), the row count array will be of size R and the column count array will be of size C. In the worst case, R and C could be proportional to N (e.g., a nearly square matrix), so the dominant space usage is proportional to the dimensions of the input, leading to O(R + C) space. Simplifying, we can express this as O(N) where N is the size of one dimension of the array (either number of rows or number of columns, whichever is larger), as the dimensions can grow linearly with the input size.

Edge Cases

Case	How to Handle
Null or empty input matrix	Return an empty matrix or throw an exception, depending on requirements, to avoid null pointer exceptions.
Matrix with zero rows or zero columns	Return an empty matrix as there are no rows or columns to process.
Matrix with only one row or one column	The logic should still function correctly, calculating differences based on the single row or column.
Very large matrix dimensions (approaching memory limits)	Consider memory usage and optimize data structures to avoid out-of-memory errors.
Matrix with all 0s or all 1s	The difference matrix will consist of the same value throughout, ensure correct computation.
Matrix with skewed distribution of 0s and 1s (e.g., mostly 0s)	The difference values can be highly negative or positive, ensure no integer overflow during calculation.
Non-square matrix (rows != columns)	The solution should handle rectangular matrices correctly without assuming square dimensions.
Integer overflow during Ones - Zeros calculation for very large matrices	Use a larger data type (e.g., long) to store the difference values to prevent potential overflow.

Null or empty input matrix

How to Handle:

Return an empty matrix or throw an exception, depending on requirements, to avoid null pointer exceptions.

Matrix with zero rows or zero columns

How to Handle:

Return an empty matrix as there are no rows or columns to process.

Matrix with only one row or one column

How to Handle:

The logic should still function correctly, calculating differences based on the single row or column.

Very large matrix dimensions (approaching memory limits)

How to Handle:

Consider memory usage and optimize data structures to avoid out-of-memory errors.

Matrix with all 0s or all 1s

How to Handle:

The difference matrix will consist of the same value throughout, ensure correct computation.

Matrix with skewed distribution of 0s and 1s (e.g., mostly 0s)

How to Handle:

The difference values can be highly negative or positive, ensure no integer overflow during calculation.

Non-square matrix (rows != columns)

How to Handle:

The solution should handle rectangular matrices correctly without assuming square dimensions.

Integer overflow during Ones - Zeros calculation for very large matrices

How to Handle:

Use a larger data type (e.g., long) to store the difference values to prevent potential overflow.