algoadvance

You are tasked with building a wall using bricks of various lengths and want to minimize the number of bricks that are crossed by a straight, vertical line. The wall is represented as a list of rows, where each row is a list of integers representing the lengths of bricks in that row, from left to right.

Find the vertical line that crosses the fewest bricks. If there are multiple answers, return any one of them.

Example:

Input: 
[[1,2,2,1],
 [3,1,2],
 [1,3,2],
 [2,4],
 [3,1,2],
 [1,3,1,1]]
Output: 2

Clarifying Questions

  1. Can the vertical line be situated at the edges of bricks?
    • Yes, the vertical line can be drawn at the edges of bricks.
  2. Can there be bricks of length zero?
    • No, all bricks have positive lengths.
  3. What to return if multiple lines cross the same number of bricks?
    • Return any one such line position.
  4. Is it possible for input lists (rows) to be empty?
    • No, each row will have at least one brick.

With these clarifications in mind, let’s proceed to devise a strategy for the solution.

Strategy

The key idea here is to find the vertical line that crosses the fewest number of bricks. If the line passes through the edge of bricks between columns, it does not count as crossing through the bricks. We should:

  1. Calculate the prefix sums for each row.
  2. Track the frequency of each prefix sum using a dictionary.
  3. The position where the sum is maximal is where the fewest bricks will be crossed.

Code

Here’s the Python code that implements the strategy:

def leastBricks(wall):
    from collections import defaultdict

    # Dictionary to count the edge positions and their occurrences
    edge_counts = defaultdict(int)

    # Calculate and store edge positions except the wall's total width
    for row in wall:
        edge_position = 0
        for brick in row[:-1]:  # exclude the last brick
            edge_position += brick
            edge_counts[edge_position] += 1

    # Find the position that has the maximum edges (minimum bricks crossed)
    max_edges = max(edge_counts.values(), default=0)

    # Number of rows minus max_edges gives us the minimum bricks to be crossed
    return len(wall) - max_edges

# Example usage
wall = [[1,2,2,1], [3,1,2], [1,3,2], [2,4], [3,1,2], [1,3,1,1]]
print(leastBricks(wall))  # Output: 2

Time Complexity

The time complexity for this algorithm is O(n * m), where n is the number of rows in the wall and m is the average number of bricks per row. This results from iterating over each brick in each row once.

The space complexity is O(w), where w is the width of the wall since we store edge positions, which is proportional to the width.

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