Solution Approach

The implementation follows the binary enumeration strategy to explore all possible column selections:

Step 1: Convert Rows to Bitmasks

First, we transform each row of the matrix into a binary number. For each row, we create a bitmask where bit j is set to 1 if matrix[i][j] == 1:

rows = [] for row in matrix:  mask = reduce(or_, (1 << j for j, x in enumerate(row) if x), 0)  rows.append(mask)

Here, 1 << j creates a number with only the j-th bit set. The reduce(or_, ...) combines all these bits using the OR operation. For example, if row is [1, 0, 1, 0], we get:

• Position 0 has 1: contribute 1 << 0 = 0001
• Position 2 has 1: contribute 1 << 2 = 0100
• Combined with OR: 0001 | 0100 = 0101 (binary) = 5 (decimal)

Step 2: Enumerate All Column Selections

We iterate through all possible column selection schemes from 0 to 2^n - 1, where n is the number of columns:

for mask in range(1 << len(matrix[0])):

Each number in this range represents a different way to select columns. For instance, with 4 columns:

• mask = 5 (binary 0101) means selecting columns 0 and 2
• mask = 3 (binary 0011) means selecting columns 0 and 1

Step 3: Filter Valid Selections

We only consider selections with exactly numSelect columns:

if mask.bit_count() != numSelect:  continue

The bit_count() method counts the number of 1 bits in the mask, which corresponds to the number of selected columns.

Step 4: Count Covered Rows

For each valid column selection, we count how many rows are covered:

t = sum((x & mask) == x for x in rows)

The expression (x & mask) == x checks if row x is covered. The AND operation x & mask keeps only the bits where both the row has a 1 AND we've selected that column. If this equals the original row x, it means all required columns for that row were selected.

Step 5: Track Maximum Coverage

We maintain the maximum number of covered rows across all valid selections:

ans = max(ans, t)

The time complexity is O(2^n × m) where n is the number of columns and m is the number of rows, since we check all 2^n possible column selections and for each, we verify coverage for all m rows.

Example Walkthrough

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

Given:

• Matrix: [[0,0,0], [1,0,1], [0,1,1], [0,0,1]]
• numSelect: 2 (we must select exactly 2 columns)

Step 1: Convert Rows to Bitmasks

Let's convert each row to its bitmask representation:

• Row 0: [0,0,0] → No 1s → bitmask = 000 (binary) = 0 (decimal)
• Row 1: [1,0,1] → 1s at positions 0 and 2 → bitmask = 101 (binary) = 5 (decimal)
• Row 2: [0,1,1] → 1s at positions 1 and 2 → bitmask = 110 (binary) = 6 (decimal)
• Row 3: [0,0,1] → 1 at position 2 → bitmask = 100 (binary) = 4 (decimal)

So our rows array becomes: [0, 5, 6, 4]

Step 2-3: Enumerate Valid Column Selections

With 3 columns total, we check masks from 0 to 7 (binary 111). We only keep those with exactly 2 bits set:

• mask = 3 (binary 011) → columns 0 and 1 selected ✓
• mask = 5 (binary 101) → columns 0 and 2 selected ✓
• mask = 6 (binary 110) → columns 1 and 2 selected ✓

Step 4: Count Covered Rows for Each Selection

Let's check each valid selection:

Selection 1: Columns 0 and 1 (mask = 011 = 3)

• Row 0 (mask 000): 000 & 011 = 000, equals row mask? Yes ✓ (covered)
• Row 1 (mask 101): 101 & 011 = 001, equals row mask? No ✗ (not covered)
• Row 2 (mask 110): 110 & 011 = 010, equals row mask? No ✗ (not covered)
• Row 3 (mask 100): 100 & 011 = 000, equals row mask? No ✗ (not covered)
• Total covered: 1 row

Selection 2: Columns 0 and 2 (mask = 101 = 5)

• Row 0 (mask 000): 000 & 101 = 000, equals row mask? Yes ✓ (covered)
• Row 1 (mask 101): 101 & 101 = 101, equals row mask? Yes ✓ (covered)
• Row 2 (mask 110): 110 & 101 = 100, equals row mask? No ✗ (not covered)
• Row 3 (mask 100): 100 & 101 = 100, equals row mask? Yes ✓ (covered)
• Total covered: 3 rows

Selection 3: Columns 1 and 2 (mask = 110 = 6)

• Row 0 (mask 000): 000 & 110 = 000, equals row mask? Yes ✓ (covered)
• Row 1 (mask 101): 101 & 110 = 100, equals row mask? No ✗ (not covered)
• Row 2 (mask 110): 110 & 110 = 110, equals row mask? Yes ✓ (covered)
• Row 3 (mask 100): 100 & 110 = 100, equals row mask? Yes ✓ (covered)
• Total covered: 3 rows

Step 5: Return Maximum

The maximum coverage is 3 rows, achieved by either selecting columns {0, 2} or columns {1, 2}.

Note how Row 0 with all zeros is always covered regardless of our column selection, while rows with 1s require all their 1-positions to be included in our selected columns to be covered.

Time and Space Complexity

The time complexity is O(2^n × m), where m is the number of rows and n is the number of columns in the matrix.

Breaking down the time complexity:

• The first loop iterates through m rows, and for each row, it processes n columns to create a bitmask, taking O(m × n) time.
• The second loop iterates through all possible column selections, which is 2^n possibilities (since each column can either be selected or not).
• For each of the 2^n masks, we check if it has exactly numSelect bits set using bit_count() in O(1) time.
• When a valid mask is found, we iterate through all m row masks to count covered rows, taking O(m) time.
• Therefore, the dominant term is O(2^n × m).

The space complexity is O(m), where m is the number of rows in the matrix.

Breaking down the space complexity:

• The rows list stores m integer bitmasks, one for each row, requiring O(m) space.
• The variables mask, ans, and t use constant space O(1).
• No recursive calls or additional data structures that scale with input size are used.
• Therefore, the total space complexity is O(m).

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

Common Pitfalls

1. Misunderstanding Row Coverage for All-Zero Rows

A critical pitfall is incorrectly handling rows that contain only zeros. Many developers mistakenly think they need special logic to check for all-zero rows.

Incorrect Approach:

# Wrong: Adding unnecessary special case handling covered = 0 for row, row_mask in zip(matrix, row_masks):  if all(val == 0 for val in row): # Unnecessary check  covered += 1  elif (row_mask & column_selection_mask) == row_mask:  covered += 1

Why This Happens: The problem statement explicitly mentions "The row contains no 1s at all (all cells are 0)" as a coverage condition, leading developers to think they need separate handling.

The Reality: When a row has all zeros, its bitmask is 0. The condition (0 & column_selection_mask) == 0 is always True regardless of which columns are selected, automatically handling this case.

Correct Approach:

# The existing code already handles all-zero rows correctly covered_rows_count = sum(  (row_mask & column_selection_mask) == row_mask  for row_mask in row_masks )

2. Off-by-One Error in Column Indexing

Another common mistake occurs when manually creating bitmasks without using enumerate properly, leading to incorrect bit positions.

Incorrect Approach:

# Wrong: Using 1-based indexing or incorrect bit shifting row_mask = 0 for j in range(len(row)):  if row[j] == 1:  row_mask |= (1 << (j + 1)) # Wrong: shifts by j+1 instead of j

Correct Approach:

# Use enumerate to ensure correct 0-based indexing row_mask = reduce(or_, (1 << col_idx for col_idx, value in enumerate(row) if value == 1), 0)

3. Integer Overflow Concerns (Language-Specific)

In languages with fixed-size integers, having too many columns could cause overflow issues. While Python handles arbitrary-precision integers automatically, this could be problematic in other languages.

Potential Issue in Other Languages:

• With 32 columns in C++ using int, the bitmask 1 << 31 might cause undefined behavior
• Solution: Use appropriate data types like unsigned long long or uint64_t

Python Advantage:

# Python handles large integers automatically for column_selection_mask in range(1 << num_columns): # Works even if num_columns > 32

4. Inefficient Bit Counting

Using manual bit counting instead of built-in methods can lead to both performance issues and potential bugs.

Inefficient/Error-Prone Approach:

# Manual bit counting - slower and more error-prone def count_bits(n):  count = 0  while n:  count += n & 1  n >>= 1  return count  if count_bits(column_selection_mask) != numSelect:  continue

Optimal Approach:

# Use built-in bit_count() method (Python 3.10+) if column_selection_mask.bit_count() != numSelect:  continue  # For older Python versions, use bin().count('1') if bin(column_selection_mask).count('1') != numSelect:  continue

These pitfalls highlight the importance of understanding how bitmask operations inherently handle edge cases and leveraging built-in language features for clarity and efficiency.