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This approach leverages the concept of 'prefix sum' where we keep a running sum across each row to determine the edges of bricks. We use a HashMap to count occurrences of these edges as potential positions to draw a line that crosses the minimum number of bricks.
Time Complexity: O(m), where m is the total number of bricks. We iterate over each brick except the last one for every row resulting in a time proportional to the number of bricks.
Space Complexity: O(width), where width represents the total width of rows. Edge positions are stored in the map, and their count can scale with the width.
1from collections import defaultdict
2
3def least_bricks(wall):
4 edge_count = defaultdict(int)
5 for row in wall:
6 prefix_sum = 0
7 for brick in row[:-1]: # Exclude the last edge in each row
8 prefix_sum += brick
9 edge_count[prefix_sum] += 1
10 if not edge_count:
11 return len(wall)
12 max_edges = max(edge_count.values())
13 return len(wall) - max_edgesThe provided Python solution employs a defaultdict to accumulate the frequency of edge positions encountered while computing the prefix sum for each brick within the rows. The goal is to find a position with the maximum edge count because this minimizes the number of bricks cut, given that the line will pass through these edges.
This alternative approach introduces a dynamic programming solution where instead of counting edges, we explore all positions by keeping track of the position sum as a state. The goal is to dynamically compute and avoid unnecessary recomputation of the edge frequencies.
Time Complexity: O(m), where m is the total number of bricks. This involves iterating through most of the bricks.
Space Complexity: O(width), where width represents the total width of rows. The storage complexity of the HashMap is proportional to how many distinct prefix sums are processed.
1import java.util.*;
2
3public class BrickWallThe core idea of this approach is to find the vertical line that crosses the least number of bricks by looking at the position where the most number of brick edges align vertically. We can achieve this by computing prefix sums for each row, except for the last brick in each row. A hashmap (or dictionary) is used to keep track of how often each prefix sum occurs.
Time Complexity: O(n * m), where n is the number of rows and m is the average number of bricks per row.
Space Complexity: O(n * m), due to the auxiliary data structure holding prefix sums.
1
This Java solution exemplifies a similar logic to the Python solution but makes use of Java collections, such as HashMap, to track and update edge counts as the prefix sums are computed. This helps find the cutting position that crosses the fewest bricks.
In this Java solution, we utilize a HashMap to store the prefix sums and their counts, ensuring that we can find the maximum number of edges aligned across the rows efficiently.