2397. Maximum Rows Covered by Columns

Problem Description

The problem provides us with a binary matrix where rows represent different items, and columns represent attributes that are either present (1) or absent (0). We are also given a number numSelect which indicates how many columns we can choose.

A row is said to be covered by the selection if for each "1" in the row, the corresponding column is part of the chosen columns, or if the row does not contain any "1"s. The goal is to select numSelect columns in such a way that the maximum number of rows are covered.

The challenge is to examine all possible combinations of columns that can be selected and then determine which combination covers the most rows.

Intuition

The intuition behind the solution is to use bit manipulation to represent the presence or absence of columns in a more computationally efficient manner. Each row of the matrix can be represented by a bitmask where the bit is set to 1 if the corresponding column contains a 1. Selecting numSelect columns is equivalent to creating a bitmask with numSelect number of 1s, representing the columns being selected.

The problem is then reduced to iterating over all possible combinations of selected columns (which are represented by bitmasks with exactly numSelect 1s). For each combination, we determine how many rows are covered by using a logical "AND" operation. A row is covered if, after the "AND" operation between the row's bitmask and the chosen columns' bitmask, the result is equal to the row's bitmask.

To count the number of 1s in a bitmask (or check if a bitmask has exactly numSelect 1s), the bit_count() function in Python is utilized. The max function is then used to keep track of the maximum number of rows that have been covered so far.

Through this approach using bitmasks and iteration, we can find the optimal selection of columns without explicitly checking each element of the subset in the original matrix, saving both time and space, and providing an optimized solution to the problem.

Learn more about Backtracking patterns.

Solution Approach

The solution approach relies on bit manipulation and enumeration. Let's walk through the implementation as provided in the reference solution with particular emphasis on the algorithms, data structures, and patterns used:

1. We start by representing each row as a bitmask. This is done by iterating over each row in the original matrix and for every "1" encountered, a corresponding bit in the mask is set. This uses list comprehension along with the reduce function from functools and the or_ bitwise operator from operator.

   1rows = []
   2for row in matrix:
   3    mask = reduce(or_, (1 << j for j, x in enumerate(row) if x), 0)
   4    rows.append(mask)

2. Next, we want to enumerate all possible combinations of columns we can choose. As each column can be represented by a bit in a bitmask, we iterate over the range 0 to 2^n (where n is the number of columns), using a for loop. Each number in this range corresponds to a potential combination of columns, with the bit at position j representing column j.

   1for mask in range(1 << len(matrix[0])):
   2    if mask.bit_count() != numSelect:
   3        continue
   4    # ... Check coverage for this combination

3. Within this loop, we use the bit_count method on the bitmask to check whether the number of columns chosen (numSelect) is equal to the number of 1s in the bitmask. If not, we skip to the next iteration:

   1if mask.bit_count() != numSelect:
   2    continue

4. For each valid combination of selected columns, we check how many rows are covered. This is done with a neat one-liner list comprehension by iterating over each row's bitmask and checking if the row's bitmask AND the selected column bitmask equals the row's bitmask, meaning the row is covered:

   1t = sum((x & mask) == x for x in rows)

5. Finally, we update our answer with the maximum value between the current answer and the number of covered rows for this column selection:

   1ans = max(ans, t)

6. Return the maximum number of rows covered after examining all combinations:

   1return ans

The overall pattern is an exhaustive search where we iterate over all possible selections of columns and, for each selection, we count how many rows are covered. This brute-force approach is made tractable by using bitmasks to efficiently represent and compare sets of columns and rows.

By compactly representing columns and rows as bits, we avoid dealing with actual column elements and row elements directly, which reduces computational complexity. However, since the number of possible combinations is 2^n, this approach is still exponential in time complexity, but it's practical for small input sizes where n (the number of columns) is not too large.

The key data structures here:

• A list rows that holds the bitmask representation of each row.
• An integer mask that holds the bitmask representation of each possible column selection.
• An integer ans that keeps track of the maximum number of rows that can be covered.

Example Walkthrough

Let's illustrate the solution approach with a small example. Consider a binary matrix matrix and numSelect as follows:

1matrix = [
2    [1, 0, 0],
3    [0, 1, 1],
4    [1, 1, 0]
5]
6numSelect = 2

Our task is to choose 2 columns that cover the maximum number of rows based on the given rules.

1. First, we convert each row of the matrix to a bitmask representation, where each "1" present in the row corresponds to a set bit in the bitmask:

   • Row 1 bitmask = 100 (in binary) = 4 (in decimal)
   • Row 2 bitmask = 011 (in binary) = 3 (in decimal)
   • Row 3 bitmask = 110 (in binary) = 6 (in decimal)

2. The list of bitmasks corresponding to the rows will be rows = [4, 3, 6].

3. Next, we need to generate all possible combinations of the columns that we can select and check if they cover the rows. Since numSelect = 2, we are looking for bitmasks with exactly 2 set bits. Our matrix has 3 columns, so we iterate over the range 0 to 2^3 or 0 to 8 in decimal.

4. The possible column combinations with 2 set bits are 110 (6 in decimal), 101 (5 in decimal) and 011 (3 in decimal). These correspond to selecting columns [1, 2], [1, 3], and [2, 3], respectively.

5. Now, check each valid column bitmask against all the row bitmasks. For each combination:

   • Consider mask = 6 (selecting columns [1, 2]):

     • Row 1 (4): (4 & 6) == 4 is True (covered)
     • Row 2 (3): (3 & 6) == 3 is True (covered)
     • Row 3 (6): (6 & 6) == 6 is True (covered)

     This selection covers all 3 rows.

   • Consider mask = 5 (selecting columns [1, 3]):

     • Row 1 (4): (4 & 5) == 4 is True (covered)
     • Row 2 (3): (3 & 5) == 1 is False (not covered)
     • Row 3 (6): (6 & 5) == 4 is False (not covered)

     This selection covers 1 row.

   • Consider mask = 3 (selecting columns [2, 3]):

     • Row 1 (4): (4 & 3) == 0 is False (not covered)
     • Row 2 (3): (3 & 3) == 3 is True (covered)
     • Row 3 (6): (6 & 3) == 2 is False (not covered)

     This selection covers 1 row.

6. The maximum number of rows covered by any selection is 3 (from the bitmask 6, which represents selecting columns [1, 2]).

7. Therefore, the output for this example would be 3 as it is the maximum number of rows that can be covered by selecting 2 columns.

Solution Implementation

Time and Space Complexity

The time complexity of the given code snippet primarily arises from the nested loops. The number of columns in the matrix is denoted as C = len(matrix[0]) and the number of rows is R = len(matrix). The for loop through every row has a complexity of O(R), and within that loop, it iterates through all columns, incurring a complexity of O(C) resulting in a total of O(R * C) for this segment.

Additionally, the code iterates through every possible combination of columns (1 << C possibilities), with a complexity of O(2^C). Within that loop, for every mask, we iterate over all the rows to perform bitmask comparisons, resulting in an O(R) complexity for this part.

Combining these, the overall time complexity is O(R * C + R * 2^C), which simplifies to O(R * (C + 2^C)).

The space complexity of this code is O(R). This arises due to the additional rows list storing a mask for each row. Each mask is an integer and there are R masks to store. The other variables used in the code are of constant size and therefore do not significantly add to the complexity.

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