#554

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Brick Wall

Solve the Brick Wall problem using array scanning and hash lookups to minimize crossed bricks from a vertical line.

Problem Statement

You are given a rectangular brick wall with several rows, each containing bricks of various widths. The height of each brick is the same, but the widths differ. The total width of each row is constant. You need to find a vertical line that crosses the fewest bricks in total by avoiding the edges of the bricks.

Given a 2D array wall where each subarray represents a row of the wall with brick widths, your task is to determine the minimum number of bricks that would be crossed by the vertical line. The line should be placed such that it does not pass along the vertical edges of the bricks.

Examples

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 details omitted.

Example 2

Input: wall = [[1],[1],[1]]

Output: 3

Example details omitted.

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

Solution Approach

Array Scanning and Hash Lookup

To minimize the number of crossed bricks, we can track the positions of brick edges. For each row in the wall, calculate the cumulative sum of brick widths and store these positions in a hash map. The most frequent cumulative position indicates where the vertical line should be placed, as it will cross the fewest bricks.

Avoiding Vertical Edges

The key insight is that drawing the vertical line near the edges of bricks is inefficient. By focusing on the positions between bricks, we ensure that fewer bricks are crossed, which can be achieved by leveraging hash table lookups for efficiency.

Optimal Complexity

The solution should use a hash table to track cumulative sums and their frequencies. This allows us to determine the most optimal position for the line with minimal time complexity, avoiding a brute force approach.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n * m), where n is the number of rows and m is the average number of bricks per row. Space complexity is O(m) for storing the cumulative sums of brick edges in the hash map.

What Interviewers Usually Probe

Look for familiarity with array scanning and hash table usage.
Candidates should demonstrate understanding of how to efficiently track cumulative sums and minimize intersections.
Watch for performance considerations when handling large inputs, as the problem constraints require optimal space and time solutions.

Common Pitfalls or Variants

Common pitfalls

Overlooking edge cases where the line goes along the vertical edge of the wall, which leads to 0 crossed bricks.
Inefficient solution using brute force rather than hash map lookups for cumulative positions.
Incorrectly handling cases where all bricks are aligned in a single column or row, leading to incorrect results.

Follow-up variants

Modify the problem by limiting the number of rows or columns to test edge cases for performance.
Allow for non-rectangular walls where row widths may differ, increasing complexity in cumulative tracking.
Change the problem to focus on drawing a horizontal line instead, requiring a different scanning pattern and data structure.

FAQ

What is the primary pattern used to solve the Brick Wall problem?

The problem is typically solved using array scanning combined with hash table lookups to track cumulative brick edges and minimize the number of crossed bricks.

How do hash tables help in solving the Brick Wall problem?

Hash tables allow us to efficiently track the positions of brick edges across rows, helping us find the most optimal vertical line with minimal crossings.

Can the Brick Wall problem be solved using brute force?

While a brute force solution is possible, it is inefficient and doesn't scale well. The optimal solution uses hash tables to avoid excessive computations.

What are the time and space complexities of the Brick Wall solution?

The time complexity is O(n * m), where n is the number of rows and m is the average number of bricks per row. Space complexity is O(m).

What are some common mistakes when solving the Brick Wall problem?

Common mistakes include inefficient brute force methods, mishandling edge cases where the line is at the vertical edge, and incorrect brick alignment handling.

terminal

Solution

Solution 1: Hash Table + Prefix Sum

We can use a hash table $\textit{cnt}$ to record the prefix sum of each row except for the last brick. The key is the value of the prefix sum, and the value is the number of times the prefix sum appears.

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class Solution:
    def leastBricks(self, wall: List[List[int]]) -> int:
        cnt = Counter()
        for row in wall:
            s = 0
            for x in row[:-1]:
                s += x
                cnt[s] += 1
        return len(wall) - max(cnt.values(), default=0)

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