Problem Description

You have a binary matrix grid of size m x n containing only 0s and 1s. You can perform operations on this matrix where each operation consists of:

• Choosing any single row and flipping all values in that row (changing all 0s to 1s and all 1s to 0s), OR
• Choosing any single column and flipping all values in that column

Your goal is to determine if it's possible to make all elements in the matrix become 0 (remove all 1s) by applying any number of these operations.

Return true if you can remove all 1s from the grid, and false if it's impossible.

For example, if you have a 2x2 matrix:

1 0
0 1

You could flip row 0 to get:

0 1
0 1

Then flip column 1 to get:

0 0
0 0

So this would return true.

The key insight is that the order of operations doesn't matter (flipping is commutative), and each row/column can be flipped at most once (flipping twice returns to original state). The solution checks if all rows can be made identical through flipping - if they can, then we can make them all zeros by flipping appropriate columns.

Intuition

Let's think about what happens when we perform these flip operations. A crucial observation is that flipping is reversible - if we flip a row or column twice, we get back to the original state. This means each row and column will be flipped either 0 times or 1 time in any optimal solution.

Now, consider what it means to remove all 1s from the grid. After all operations, every position must be 0. Let's think about any two rows in the final state - they must both be all zeros, meaning they're identical.

Working backwards, if two rows end up identical (all zeros) after operations, what must have been true about them originally? Since we can only flip entire rows or columns:

• If we flip a column, it affects both rows in the same way
• If we flip one row but not the other, they become "opposite" to each other

This leads to a key insight: two rows can only end up identical if they were originally either identical or exactly opposite (one is the bitwise complement of the other).

Therefore, all rows in the original grid must fall into at most two groups:

1. Rows that are identical to some reference row
2. Rows that are the exact complement of that reference row

We can normalize this by choosing the first row as our reference. For any other row:

• If it starts with the same value as the first row's first element, it should be identical to the first row
• If it starts with a different value, it should be the exact complement of the first row

The solution checks if this pattern holds by converting all rows to the same "form" (flipping complementary rows) and verifying they're all identical. If they are, we can make them all zeros; if not, it's impossible.

Learn more about Math patterns.

Solution Approach

The implementation uses a set-based approach to verify if all rows can be normalized to the same pattern:

1. Initialize a set s to store unique row patterns after normalization.

2. Process each row in the grid:

   • Check if the first element of the current row equals the first element of the first row (row[0] == grid[0][0])
   • If yes, keep the row as-is: t = tuple(row)
   • If no, flip all bits in the row: t = tuple(x ^ 1 for x in row)
   • This normalization ensures that all rows that can be made identical through flipping will have the same representation

3. Add the normalized row to the set: Since sets only store unique elements, if all rows can be made identical, the set will contain exactly one element.

4. Check the set size: Return len(s) == 1

   • If the set has only one element, all rows can be transformed to be identical, meaning we can remove all 1s
   • If the set has more than one element, it's impossible to make all rows identical

The algorithm leverages the XOR operation (x ^ 1) to flip bits efficiently, where 0 ^ 1 = 1 and 1 ^ 1 = 0.

The time complexity is O(m × n) where we iterate through all elements once. The space complexity is O(n) in the best case (all rows identical) or O(m × n) in the worst case (all rows different).

The key insight is using the first row as a reference point and normalizing all other rows relative to it, which elegantly handles the problem of determining whether rows can be made identical through any combination of row and column flips.

Example Walkthrough

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

Consider this 3x3 matrix:

1 0 1
0 1 0
1 0 1

Step 1: Initialize our set s We'll use this set to store normalized row patterns.

Step 2: Process each row

Row 0: [1, 0, 1]

• This is our first row, so it becomes our reference
• Check: row[0] == grid[0][0] → 1 == 1 → True
• Since it matches, we keep the row as-is
• Add to set: s = {(1, 0, 1)}

Row 1: [0, 1, 0]

• Check: row[0] == grid[0][0] → 0 == 1 → False
• Since it doesn't match, we flip all bits: 0→1, 1→0, 0→1
• Flipped row: [1, 0, 1]
• Add to set: s = {(1, 0, 1)} (no change, already exists)

Row 2: [1, 0, 1]

• Check: row[0] == grid[0][0] → 1 == 1 → True
• Since it matches, we keep the row as-is
• Add to set: s = {(1, 0, 1)} (no change, already exists)

Step 3: Check set size

• len(s) == 1 → True
• All rows normalized to the same pattern (1, 0, 1)

Result: Return true

This means we can remove all 1s. Here's how:

1. Flip row 1: Matrix becomes [[1,0,1], [1,0,1], [1,0,1]]
2. Flip column 0: Matrix becomes [[0,0,1], [0,0,1], [0,0,1]]
3. Flip column 2: Matrix becomes [[0,0,0], [0,0,0], [0,0,0]]

Counter-example where it's impossible:

Consider this matrix:

1 0 1
0 1 0
0 0 1

Processing:

• Row 0: [1,0,1] → keep as-is → (1,0,1)
• Row 1: [0,1,0] → flip → (1,0,1)
• Row 2: [0,0,1] → flip → (1,1,0)

Set contains: {(1,0,1), (1,1,0)} → len(s) == 2 → Return false

The third row, when normalized, doesn't match the pattern of the first two rows, so it's impossible to make all elements zero.

Solution Implementation

Time and Space Complexity

Time Complexity: O(m * n) where m is the number of rows and n is the number of columns in the grid.

• The outer loop iterates through each row in the grid: O(m) iterations
• For each row, we either create a tuple directly from the row (O(n)) or create a tuple with XOR operations (O(n))
• The condition check row[0] == grid[0][0] is O(1)
• Adding to the set is O(n) on average (since tuple hashing is O(n) for a tuple of length n)
• Overall: O(m * n)

Space Complexity: O(m * n) in the worst case.

• The set s can contain at most m tuples (one for each row)
• Each tuple has n elements
• In the worst case where all processed tuples are unique, the space used is O(m * n)
• The tuple creation also uses O(n) temporary space for each row
• Overall: O(m * n)

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

Common Pitfalls

1. Misunderstanding the Problem Statement

A common misunderstanding is thinking we need to literally "remove" rows and columns that contain all 1s. The problem actually asks to make all elements become 0 through flipping operations, not removing rows/columns. The function name removeOnes can be misleading.

Solution: Focus on the flipping operations - we can flip entire rows or columns to change 0s to 1s and vice versa.

2. Incorrect Normalization Logic

A critical pitfall is normalizing rows incorrectly. Some might try to normalize based on whether the row has more 0s or 1s, or always flip rows that start with 1:

# WRONG: Always flipping rows that start with 1
for row in grid:
    if row[0] == 1:
        normalized = tuple(x ^ 1 for x in row)
    else:
        normalized = tuple(row)

This fails because we need consistency across all rows relative to a reference point.

Solution: Always compare each row's first element with the first row's first element to maintain consistency:

if current_row[0] == grid[0][0]:  # Compare with reference
    normalized = tuple(current_row)
else:
    normalized = tuple(x ^ 1 for x in current_row)

3. Forgetting to Convert to Immutable Type

Using lists directly in a set will cause a TypeError since lists are mutable and unhashable:

# WRONG: This will raise TypeError
unique_patterns = set()
unique_patterns.add(current_row)  # Lists can't be added to sets

Solution: Convert rows to tuples before adding to the set:

normalized_row = tuple(current_row)  # or tuple(x ^ 1 for x in current_row)
unique_patterns.add(normalized_row)

4. Over-complicating the Solution

Some might try to actually simulate the flipping operations or use complex mathematical approaches like Gaussian elimination over GF(2). This makes the solution unnecessarily complex.

Solution: Recognize that the order of operations doesn't matter and each row/column needs at most one flip. The simple normalization approach is sufficient.

5. Not Handling Edge Cases

Forgetting to consider edge cases like:

• Single row or single column matrices
• Matrices that are already all zeros
• Matrices where all rows are already identical

Solution: The current approach naturally handles these cases, but it's important to verify:

• A 1×n or m×1 matrix will have len(unique_patterns) == 1
• An all-zero matrix will have one pattern: (0, 0, ..., 0)
• Already identical rows will produce len(unique_patterns) == 1