How do you solve Maximum Rows Covered by Columns on LeetCode?

Maximum Rows Covered by Columns (LeetCode #2397) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^n) time complexity and O(1) space complexity. Key insight: Using bit manipulation can significantly reduce the complexity of the problem.

What is the brute force solution for Maximum Rows Covered by Columns?

The brute force approach for Maximum Rows Covered by Columns is Brute Force, with O(n^2) time complexity and O(1) space complexity. The brute-force approach involves generating all possible combinations of columns and checking how many rows can be covered by each combination. This method is straightforward but inefficient for larger matrices. The optimal Optimal Solution approach improves this to O(n * 2^n).

What algorithm or data structure is used to solve Maximum Rows Covered by Columns?

Maximum Rows Covered by Columns uses the following concepts: Array, Backtracking, Bit Manipulation, Matrix, Enumeration. The recommended patterns to study are: Bit Manipulation, Combinatorial Enumeration.

What are the interview tips for Maximum Rows Covered by Columns?

Clarify the problem and constraints before jumping into coding. Think about how to represent combinations efficiently. Test your solution with edge cases to ensure robustness.

What are common mistakes when solving Maximum Rows Covered by Columns?

Failing to account for rows that have no 1's. Not considering edge cases like all rows being covered or none.

#2397

Maximum Rows Covered by Columns

Medium

ArrayBacktrackingBit ManipulationMatrixEnumerationBit ManipulationCombinatorial Enumeration

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n^2)	O(n * 2^n)
Space	O(1)	O(1)

💡

Intuition

Time O(n * 2^n)Space O(1)

The optimal solution leverages bit manipulation to efficiently represent and check combinations of columns. This reduces the complexity significantly by avoiding the need to generate all combinations explicitly.

⚙️

Algorithm

3 steps

1Step 1: Use bit masking to represent selected columns.
2Step 2: For each possible combination of columns (from 0 to 2^n - 1), check if it covers the maximum rows.
3Step 3: Count the number of rows covered for each combination and keep track of the maximum.

solution.py11 lines

1def maxRowsCovered(matrix, numSelect):
2    m, n = len(matrix), len(matrix[0])
3    max_covered = 0
4    for mask in range(1 << n):
5        if bin(mask).count('1') == numSelect:
6            covered = 0
7            for row in matrix:
8                if all(row[j] == 0 for j in range(n) if (mask & (1 << j)) == 0) or all(row[j] == 1 for j in range(n) if (mask & (1 << j)) != 0):
9                    covered += 1
10            max_covered = max(max_covered, covered)
11    return max_covered

ℹ

Complexity note: The complexity comes from iterating through all possible combinations of columns (2^n) and checking each row (n) for coverage. This is more efficient than the brute force approach as it avoids generating combinations explicitly.

1Using bit manipulation can significantly reduce the complexity of the problem.
2Understanding how to represent combinations efficiently is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.