Brick Wall Leetcode Problem 554 [Python Solution]

In this blog post, we'll tackle the Brick Wall problem from LeetCode, specifically problem number 554. This problem falls under the category of arrays and hashing and is considered to have a medium level of difficulty.

The goal is to find an efficient solution to minimize the number of bricks cut when drawing a vertical line through a rectangular brick wall.

We will explore the problem, discuss constraints, present both a brute-force and an efficient approach, provide Python code solutions, analyze time and space complexity, and explain the reasoning behind our efficient approach.

Problem Overview

Imagine you are facing a rectangular brick wall consisting of multiple rows of bricks.

Each row of bricks has the same height (one unit), but the individual bricks within a row may vary in width.

The total width of each row is identical.

Your task is to draw a vertical line from the top to the bottom of the wall in such a way that it crosses the fewest number of bricks.

However, if the line goes along the edge of a brick, it is not considered as crossed.

You cannot draw the line solely along one of the two vertical edges of the wall since that would cross no bricks.

Given a 2D array called wall, which contains information about the wall's structure, you need to determine the minimum number of bricks that will be crossed when drawing the 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 is the number of rows in the wall.
• 1 <= n <= 10,000
• Each row wall[i] contains between 1 and 10,000 bricks.
• The total sum of brick widths in each row does not exceed 20,000.
• The width of each brick, wall[i][j], is in the range of 1 to 2,147,483,647.

Understanding the Constraints

Before delving into the problem's solution, let's take a moment to understand the constraints.

1. The number of rows in the wall, denoted by n, can range from 1 to 10,000. This means we might have to deal with a large number of rows, and our solution should be efficient in terms of time complexity.

2. Each row, represented as wall[i], contains between 1 and 10,000 bricks.

While the minimum number of bricks in a row is 1, the maximum can be quite large.

This implies that we might need to iterate through many bricks when calculating the minimum number of crosses.

1. The total sum of brick widths in each row does not exceed 20,000. This constraint indicates that the total width of bricks in a row remains within reasonable bounds.

It also suggests that the input size of each row is not extremely large.

1. The width of each brick, wall[i][j], is in the range of 1 to 2,147,483,647. The width of individual bricks can be very large, so our solution should efficiently handle these large values and not cause integer overflow.

Brute-force Approach

Let's start by discussing a brute-force approach to solving this problem.

The idea is to try every possible position for the vertical line within the wall, excluding the edges, and count how many bricks are crossed at each position.

The position that results in the fewest crosses will be our answer.

Python Code for Brute-force Approach

def leastBricks(wall):
    countGap = {0: 0}  # {Position: Gap count}

    for row in wall:
        total = 0  # Position
        for brick in row[:-1]:  # Exclude the last brick's width
            total += brick
            countGap[total] = 1 + countGap.get(total, 0)

    return len(wall) - max(countGap.values())

In this Python code, we create a countGap dictionary to keep track of the gaps' counts at different positions.

We iterate through each row of the wall and calculate the position of the line as we move from left to right within each row.

We use countGap to store the number of gaps we encounter at each position.

The countGap dictionary is structured with position as the key and gap count as the value.

We start with a position of 0 (the leftmost edge of the wall) and initialize the gap count as 0. As we encounter gaps between bricks within a row, we update the gap count for the corresponding position.

If the position does not exist in the dictionary, we add it with a gap count of 1. Finally, we return the difference between the total number of rows in the wall and the maximum gap count in countGap.

This gives us the minimum number of bricks that need to be crossed.

Efficient Approach with Hash Map

While the brute-force approach is a straightforward way to solve the problem, it's not the most efficient solution.

We can optimize our approach using a hash map.

The key insight is to find the positions where the gaps between bricks are most frequently occurring, as these are the optimal positions for drawing the line.

Python Code for Efficient Approach

def leastBricks(wall):
    countGap = {0: 0}  # {Position: Gap count}

    for row in wall:
        total = 0  # Position
        for brick in row[:-1]:  # Exclude the last brick's width
            total += brick
            countGap[total] = 1 + countGap.get(total, 0)

    return len(wall) - max(countGap.values())

This efficient approach is essentially the same as the brute-force one but with a crucial optimization.

We still use the countGap dictionary to store gap counts at different positions.

The critical insight is that the position with the highest gap count in this dictionary will correspond to the optimal position for drawing the line.

By finding the maximum gap count, we can determine where the line should be drawn to minimize the number of bricks crossed.

Our goal is to maximize the number of gaps crossed and, consequently, minimize the number of bricks crossed.

The final result is calculated by subtracting the maximum gap count from the total number of rows in the wall.

Time and Space Complexity

Let's analyze the time and space complexity of our efficient solution.

Time Complexity:

1. Iterating through each row in the wall: O(n), where n is the number of rows.
2. In each row, iterating through the bricks (excluding the last one): O(m), where m is the maximum number of bricks in a row.
3. Updating the countGap dictionary: O(m).
4. Finding the maximum gap count: O(m).

The overall time complexity is O(n * m), where n is the number of rows, and m is the maximum number of bricks in a row.

Space Complexity:

1. countGap dictionary: O(m) space is required to store positions

   and their corresponding gap counts.

2. Other variables and data structures: O(1) space.

The overall space complexity is O(m).

Reasoning Behind Our Efficient Approach

The key to solving this problem efficiently is to identify the positions where gaps between bricks are most frequent.

By focusing on these positions, we can minimize the number of bricks crossed when drawing the line.

Here's the reasoning behind our efficient approach:

1. We start by initializing a dictionary called countGap to keep track of gap counts at different positions.

The position represents the cumulative width of bricks from the leftmost edge of the wall.

1. We iterate through each row of the wall, and for each row, we traverse the bricks while excluding the last one.

The reason for excluding the last brick is that we cannot draw the line along the rightmost edge of the wall, and there are no gaps beyond that edge.

1. As we move from left to right within a row, we calculate the cumulative position.

If a gap between two bricks is encountered, we increment the gap count for that position in the countGap dictionary.

1. After processing all rows, the countGap dictionary contains the gap counts for various positions.

To find the optimal position for drawing the line, we look for the position with the highest gap count.

1. The reason for selecting the position with the maximum gap count is that drawing the line at this position maximizes the number of gaps crossed, which, in turn, minimizes the number of bricks crossed.

2. We compute the final result by subtracting the maximum gap count from the total number of rows in the wall.

This gives us the minimum number of bricks that need to be crossed when drawing the line.

By identifying the positions with the most gaps, we efficiently solve the problem without having to consider every possible position, resulting in a faster and more optimized solution.

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Conclusion

In this blog post, we explored the Brick Wall problem from LeetCode, which falls under the category of arrays and hashing.

We discussed the problem's statement and constraints, and we presented both a brute-force and an efficient approach to finding the minimum number of bricks to cross when drawing a vertical line through a brick wall.

Our efficient approach, based on using a hash map to track gap counts at different positions, allows us to solve the problem in a more optimized way.

By identifying the position with the maximum gap count, we determine the optimal line placement that minimizes the number of bricks crossed.

The time complexity of our solution is O(n * m), where n is the number of rows and m is the maximum number of bricks in a row.

The space complexity is O(m).

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You can also access the original problem on LeetCode by following this link.

Thank you for joining us in solving the Brick Wall problem, and we look forward to sharing more coding challenges with you in the future.

Happy coding!