Introduction

In this blog post, we will explore how to solve the LeetCode problem "554" 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

There is a rectangular brick wall in front of you with n rows of bricks. The ith row has some number of bricks each of the same height (i.e., one unit) but they can be of different widths. The total width of each row is the same. Draw a vertical line from the top to the bottom and cross the least bricks. If your line goes through the edge of a brick, then the brick is not considered as crossed. You cannot draw a line just along one of the two vertical edges of the wall, in which case the line will obviously cross no bricks. Given the 2D array wall that contains the information about the wall, return the minimum number of crossed bricks after drawing such a vertical line. Example 1: Input: wall = [[1,2,2,1],[3,1,2],[1,3,2],[2,4],[3,1,2],[1,3,1,1]] Output: 2 Example 2: Input: wall = [[1],[1],[1]] Output: 3 Constraints: n == wall.length 1 <= n <= 104 1 <= wall[i].length <= 104 1 <= sum(wall[i].length) <= 2 * 104 sum(wall[i]) is the same for each row i. 1 <= wall[i][j] <= 231 - 1

Explanation

Here's a breakdown of the approach, complexity, and Python code:

• Core Idea: The optimal vertical line will pass through the most common "gap" location across all rows. We can find these common gaps by iterating through each row, calculating the cumulative width at each brick edge (excluding the very last edge of the wall in each row), and storing the frequency of each cumulative width using a hash map.
• Optimization: After counting gap frequencies, the minimum number of crossed bricks equals the total number of rows minus the maximum frequency of any gap. This is because, for every row where the optimal line passes through a gap, we don't cross a brick.

• Edge Case Handling: The problem statement specifies that we cannot draw a line along the edge of the wall, therefore, we don't consider the total width of the row when calculating the cumulative widths (i.e. the very last edge of the wall in each row).

• Complexity: O(N), where N is the total number of bricks in the wall, and O(M) space, where M is the number of unique edge locations.

Code

    def leastBricks(wall):
    """
    Finds the minimum number of crossed bricks after drawing a vertical line.

    Args:
        wall: A 2D list of integers representing the wall.

    Returns:
        The minimum number of crossed bricks.
    """

    gap_counts = {}
    for row in wall:
        cumulative_width = 0
        for i in range(len(row) - 1):  # Exclude the last brick in each row
            cumulative_width += row[i]
            gap_counts[cumulative_width] = gap_counts.get(cumulative_width, 0) + 1

    if not gap_counts:
        return len(wall)  # Handle the case where all rows have only one brick

    max_gap_frequency = 0
    for count in gap_counts.values():
        max_gap_frequency = max(max_gap_frequency, count)

    return len(wall) - max_gap_frequency