Crossing the least number of bricks is equivalent to crossing the most number of edges. We use a hash map to count the frequencies of the vertical edges (positions) across all rows.

1. Initialize edges = {0: 0}.
2. For each row in wall:
   • curr_pos = 0.
   • Iterate through bricks in row (except the last one):
     • curr_pos += brick_width.
     • edges[curr_pos] = edges.get(curr_pos, 0) + 1.
3. The number of rows is len(wall).
4. Max edges found is max(edges.values()).
5. Minimum crossed bricks = total_rows - max_edges.

• Time Complexity: O(N), where N is total bricks.
• Space Complexity: O(W), where W is the width of the wall (or number of unique edge positions).

def least_bricks(wall):
    edges = {0: 0} # map position to edge count
    for row in wall:
        pos = 0
        for brick in row[:-1]:
            pos += brick
            edges[pos] = edges.get(pos, 0) + 1
            
    return len(wall) - max(edges.values())