Problem Description

You are given a binary matrix grid with dimensions m x n (m rows and n columns) containing only 0s and 1s. Your task is to create a new matrix called diff with the same dimensions, where each element is calculated using a specific formula based on the counts of 1s and 0s in the corresponding row and column.

For each position [i][j] in the difference matrix, you need to:

1. Count the number of 1s in row i (call this onesRow[i])
2. Count the number of 1s in column j (call this onesCol[j])
3. Count the number of 0s in row i (call this zerosRow[i])
4. Count the number of 0s in column j (call this zerosCol[j])

Then calculate: diff[i][j] = onesRow[i] + onesCol[j] - zerosRow[i] - zerosCol[j]

The solution approach first counts the number of 1s in each row and column. Since we know the total number of elements in each row is n and in each column is m, we can derive:

• zerosRow[i] = n - onesRow[i] (total columns minus ones in that row)
• zerosCol[j] = m - onesCol[j] (total rows minus ones in that column)

The formula then simplifies to: diff[i][j] = onesRow[i] + onesCol[j] - (n - onesRow[i]) - (m - onesCol[j])

This can be further simplified to: diff[i][j] = 2 * onesRow[i] + 2 * onesCol[j] - n - m

The algorithm iterates through the grid once to count 1s in each row and column, then builds the difference matrix using these precomputed values, achieving an efficient O(m*n) time complexity.

Intuition

The key insight is recognizing that we don't need to recalculate the counts for every single cell. If we look at the formula diff[i][j] = onesRow[i] + onesCol[j] - zerosRow[i] - zerosCol[j], we notice that for any cell in row i, the values onesRow[i] and zerosRow[i] remain constant. Similarly, for any cell in column j, the values onesCol[j] and zerosCol[j] stay the same.

This observation leads us to think about preprocessing: instead of counting 1s and 0s repeatedly for each cell, we can count them once for each row and column, then reuse these values.

Another crucial realization is that we don't need to separately count both 1s and 0s. Since each row has exactly n elements and each column has exactly m elements, if we know the count of 1s, we can immediately derive the count of 0s:

• Number of 0s in row i = n - Number of 1s in row i
• Number of 0s in column j = m - Number of 1s in column j

This relationship allows us to simplify our formula. If we let r be the count of 1s in row i and c be the count of 1s in column j, then:

• onesRow[i] = r
• onesCol[j] = c
• zerosRow[i] = n - r
• zerosCol[j] = m - c

Substituting into the original formula: diff[i][j] = r + c - (n - r) - (m - c) = 2r + 2c - n - m

This transformation shows us that we only need to track the count of 1s for each row and column, making the solution both simpler and more efficient. We can compute all row and column counts in a single pass through the matrix, then construct the difference matrix using these precomputed values in a second pass.

Solution Approach

The implementation follows a two-pass approach using arrays to store precomputed counts:

Step 1: Initialize Storage Arrays We create two arrays to store the count of 1s:

• rows array of size m to store the count of 1s in each row
• cols array of size n to store the count of 1s in each column

Step 2: Count 1s in Single Pass We iterate through the entire grid once. For each cell grid[i][j]:

• If the value is 1, we increment both rows[i] and cols[j]
• This is done efficiently by adding the cell value directly: rows[i] += v and cols[j] += v

for i, row in enumerate(grid):
    for j, v in enumerate(row):
        rows[i] += v  # Add 1 if cell is 1, add 0 if cell is 0
        cols[j] += v

Step 3: Build the Difference Matrix We create a new m x n matrix diff and populate it using our precomputed counts. For each position [i][j]:

• r represents the count of 1s in row i
• c represents the count of 1s in column j
• Apply the simplified formula: diff[i][j] = r + c - (n - r) - (m - c)

diff = [[0] * n for _ in range(m)]
for i, r in enumerate(rows):
    for j, c in enumerate(cols):
        diff[i][j] = r + c - (n - r) - (m - c)

The formula r + c - (n - r) - (m - c) effectively calculates:

• r (ones in row i) + c (ones in column j)
• minus (n - r) (zeros in row i)
• minus (m - c) (zeros in column j)

Time Complexity: O(m * n) - We traverse the grid twice, once for counting and once for building the result.

Space Complexity: O(m + n) for the auxiliary arrays (not counting the output matrix).

This approach is optimal because we must examine every cell at least once to get the counts, and we must fill every cell in the output matrix, giving us a lower bound of O(m * n) time.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Given grid:

grid = [[0, 1, 1],
        [1, 0, 1],
        [0, 0, 0]]

Step 1: Count 1s in each row and column

First pass through the grid:

• Row 0: [0, 1, 1] → contains 2 ones → rows[0] = 2

• Row 1: [1, 0, 1] → contains 2 ones → rows[1] = 2

• Row 2: [0, 0, 0] → contains 0 ones → rows[2] = 0

• Column 0: [0, 1, 0] → contains 1 one → cols[0] = 1

• Column 1: [1, 0, 0] → contains 1 one → cols[1] = 1

• Column 2: [1, 1, 0] → contains 2 ones → cols[2] = 2

So we have:

• rows = [2, 2, 0]
• cols = [1, 1, 2]
• m = 3 (number of rows)
• n = 3 (number of columns)

Step 2: Calculate diff matrix using the formula

For each position [i][j], we apply: diff[i][j] = rows[i] + cols[j] - (n - rows[i]) - (m - cols[j])

Position [0][0]:

• rows[0] = 2, cols[0] = 1
• diff[0][0] = 2 + 1 - (3 - 2) - (3 - 1) = 3 - 1 - 2 = 0

Position [0][1]:

• rows[0] = 2, cols[1] = 1
• diff[0][1] = 2 + 1 - (3 - 2) - (3 - 1) = 3 - 1 - 2 = 0

Position [0][2]:

• rows[0] = 2, cols[2] = 2
• diff[0][2] = 2 + 2 - (3 - 2) - (3 - 2) = 4 - 1 - 1 = 2

Position [1][0]:

• rows[1] = 2, cols[0] = 1
• diff[1][0] = 2 + 1 - (3 - 2) - (3 - 1) = 3 - 1 - 2 = 0

Position [1][1]:

• rows[1] = 2, cols[1] = 1
• diff[1][1] = 2 + 1 - (3 - 2) - (3 - 1) = 3 - 1 - 2 = 0

Position [1][2]:

• rows[1] = 2, cols[2] = 2
• diff[1][2] = 2 + 2 - (3 - 2) - (3 - 2) = 4 - 1 - 1 = 2

Position [2][0]:

• rows[2] = 0, cols[0] = 1
• diff[2][0] = 0 + 1 - (3 - 0) - (3 - 1) = 1 - 3 - 2 = -4

Position [2][1]:

• rows[2] = 0, cols[1] = 1
• diff[2][1] = 0 + 1 - (3 - 0) - (3 - 1) = 1 - 3 - 2 = -4

Position [2][2]:

• rows[2] = 0, cols[2] = 2
• diff[2][2] = 0 + 2 - (3 - 0) - (3 - 2) = 2 - 3 - 1 = -2

Final Result:

diff = [[ 0,  0,  2],
        [ 0,  0,  2],
        [-4, -4, -2]]

This example demonstrates how we avoid redundant counting by precomputing the row and column sums once, then efficiently calculating each cell in the difference matrix using the simplified formula.

Solution Implementation

Time and Space Complexity

Time Complexity: O(m × n)

The algorithm consists of two main phases:

1. First nested loop: Iterates through all m × n elements of the grid to calculate row sums and column sums. This takes O(m × n) time.
2. Second nested loop: Iterates through all m rows and n columns to compute the difference matrix. This also takes O(m × n) time.

Since both phases are sequential (not nested with each other), the total time complexity is O(m × n) + O(m × n) = O(m × n).

Space Complexity: O(m + n) (excluding the output array)

The algorithm uses:

• rows array of size m to store row sums: O(m) space
• cols array of size n to store column sums: O(n) space
• diff matrix of size m × n for the output: O(m × n) space

Following the reference answer's convention of ignoring the space used by the answer (the diff matrix), the space complexity is O(m) + O(n) = O(m + n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrectly Calculating Zeros Count

A common mistake is trying to count zeros separately or miscalculating the relationship between ones and zeros.

Incorrect approach:

# Wrong: Counting zeros separately (inefficient and error-prone)
zeros_in_row = [0] * num_rows
zeros_in_col = [0] * num_cols
for i, row in enumerate(grid):
    for j, value in enumerate(row):
        if value == 0:
            zeros_in_row[i] += 1
            zeros_in_col[j] += 1

Correct approach:

# Right: Derive zeros from ones count
zeros_in_row[i] = num_cols - ones_in_row[i]
zeros_in_col[j] = num_rows - ones_in_col[j]

2. Confusion Between Row and Column Dimensions

Mixing up m (number of rows) and n (number of columns) when calculating zeros is a frequent error.

Incorrect:

# Wrong: Using num_rows for row zeros calculation
diff_matrix[i][j] = (
    row_ones + col_ones - 
    (num_rows - row_ones) -  # Should be num_cols!
    (num_cols - col_ones)    # Should be num_rows!
)

Correct:

# Right: Row has num_cols elements, column has num_rows elements
diff_matrix[i][j] = (
    row_ones + col_ones - 
    (num_cols - row_ones) -  # Zeros in row i
    (num_rows - col_ones)    # Zeros in column j
)

3. Modifying Input Grid Instead of Creating New Matrix

Some might try to modify the input grid in-place to save space, which changes the original data.

Incorrect:

# Wrong: Modifying input grid
for i in range(num_rows):
    for j in range(num_cols):
        grid[i][j] = calculation_result  # Destroys original data
return grid

Correct:

# Right: Create new matrix
diff_matrix = [[0] * num_cols for _ in range(num_rows)]
# ... populate diff_matrix
return diff_matrix

4. Off-by-One Errors in Formula Application

Forgetting to subtract or add certain components of the formula.

Incorrect:

# Wrong: Missing subtraction or using wrong signs
diff_matrix[i][j] = row_ones + col_ones + zeros_in_row + zeros_in_col
# Or
diff_matrix[i][j] = row_ones - col_ones - zeros_in_row - zeros_in_col

Correct:

# Right: Proper formula application
diff_matrix[i][j] = (
    ones_in_row[i] + ones_in_col[j] - 
    zeros_in_row[i] - zeros_in_col[j]
)

5. Inefficient Matrix Initialization

Using append operations instead of pre-allocating the matrix size.

Inefficient:

# Suboptimal: Building matrix row by row
diff_matrix = []
for i in range(num_rows):
    row = []
    for j in range(num_cols):
        row.append(calculated_value)
    diff_matrix.append(row)

Efficient:

# Better: Pre-allocate matrix
diff_matrix = [[0] * num_cols for _ in range(num_rows)]
# Then fill in values

Prevention Tips:

1. Always clarify what represents rows vs columns (m vs n)
2. Use descriptive variable names that indicate what they count
3. Double-check the mathematical relationship: zeros = total - ones
4. Test with simple examples where you can manually verify the calculation
5. Remember that row i has n elements and column j has m elements