Introduction

In this blog post, we will explore how to solve the LeetCode problem "2397" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an m x n binary matrix matrix and an integer numSelect. Your goal is to select exactly numSelect distinct columns from matrix such that you cover as many rows as possible. A row is considered covered if all the 1's in that row are also part of a column that you have selected. If a row does not have any 1s, it is also considered covered. More formally, let us consider selected = {c1, c2, ...., cnumSelect} as the set of columns selected by you. A row i is covered by selected if: For each cell where matrix[i][j] == 1, the column j is in selected. Or, no cell in row i has a value of 1. Return the maximum number of rows that can be covered by a set of numSelect columns. Example 1: Input: matrix = [[0,0,0],[1,0,1],[0,1,1],[0,0,1]], numSelect = 2 Output: 3 Explanation: One possible way to cover 3 rows is shown in the diagram above. We choose s = {0, 2}. - Row 0 is covered because it has no occurrences of 1. - Row 1 is covered because the columns with value 1, i.e. 0 and 2 are present in s. - Row 2 is not covered because matrix[2][1] == 1 but 1 is not present in s. - Row 3 is covered because matrix[2][2] == 1 and 2 is present in s. Thus, we can cover three rows. Note that s = {1, 2} will also cover 3 rows, but it can be shown that no more than three rows can be covered. Example 2: Input: matrix = [[1],[0]], numSelect = 1 Output: 2 Explanation: Selecting the only column will result in both rows being covered since the entire matrix is selected. Constraints: m == matrix.length n == matrix[i].length 1 <= m, n <= 12 matrix[i][j] is either 0 or 1. 1 <= numSelect <= n

Explanation

• Represent Rows as Bitmasks: Convert each row into a bitmask, where the i-th bit is set if matrix[row][i] is 1. This allows us to efficiently check if a row is covered by a set of selected columns.
  • Iterate Through Column Combinations: Generate all possible combinations of numSelect columns. For each combination, check how many rows are covered.
  • Maximize Covered Rows: Keep track of the maximum number of covered rows found so far and update it as we iterate through column combinations.

• Runtime Complexity: O(n choose numSelect m), where n is the number of columns and m is the number of rows. More precisely, it is O(n! / ((n - numSelect)! numSelect!) * m)
• Storage Complexity: O(m)

Code

    from itertools import combinations

def maximumRows(matrix, numSelect):
    m = len(matrix)
    n = len(matrix[0])
    max_covered = 0

    for col_indices in combinations(range(n), numSelect):
        covered_rows = 0
        for row in range(m):
            covered = True
            for col in range(n):
                if matrix[row][col] == 1 and col not in col_indices:
                    covered = False
                    break
            if covered:
                covered_rows += 1
        max_covered = max(max_covered, covered_rows)

    return max_covered