554. Brick Wall

LeetCode 554. Brick Wall

Description

There is a brick wall in front of you. The wall is rectangular and has several rows of bricks. The bricks have the same height but different width. You want to draw a vertical line from the top to the bottom and cross the least bricks.

The brick wall is represented by a list of rows. Each row is a list of integers representing the width of each brick in this row from left to right.

If your line go through the edge of a brick, then the brick is not considered as crossed. You need to find out how to draw the line to cross the least bricks and return the number of crossed bricks.

You cannot draw a line just along one of the two vertical edges of the wall, in which case the line will obviously cross no bricks.

Example:

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

Note:

1. The width sum of bricks in different rows are the same and won't exceed INT_MAX.

2. The number of bricks in each row is in range [1,10,000]. The height of wall is in range [1,10,000]. Total number of bricks of the wall won't exceed 20,000.

Tags

Hash Table

Solution

To obtain the least bricks, we can compute the most edges first and return len(wall) - maxEdges finally. To achieve this, we initialize a hash table whose key is the position of edge and value is the number of edges at this position. Thus, for each line, we use variable i to accumulate the width of each brick and increase that hash table pos2edges[i] by 1.

Complexity

• Time complexity: $O(mn)$

• Space complexity: $O(mn)$

Code

func leastBricks(wall [][]int) int {
	pos2edges := map[int]int{}
	for _, widths := range wall {
		var i int
		for _, w := range widths[:len(widths)-1] {
			i += w
			pos2edges[i]++
		}
	}
	var maxEdges int
	for _, e := range pos2edges {
		if maxEdges < e {
			maxEdges = e
		}
	}
	return len(wall) - maxEdges
}

Reference

1. 砖墙