# Brick Wall LeetCode Solution

Here, We see Brick Wall LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

There is a rectangular brick wall in front of you with `n` rows of bricks. The `ith` row has some number of bricks each of the same height (i.e., one unit) but they can be of different widths. The total width of each row is the same.

Draw a vertical line from the top to the bottom and cross the least bricks. If your line goes through the edge of a brick, then the brick is not considered as crossed. 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.

Given the 2D array `wall` that contains the information about the wall, return the minimum number of crossed bricks after drawing such a vertical line.

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 2:
Input: wall = [[1],[1],[1]]
Output: 3

## Brick Wall LeetCode SolutionC++

``````class Solution {
public:
int leastBricks(vector<vector<int>>& wall) {
unordered_map<int, int> edge_frequency;
int max_frequency = 0;
for(int row=0; row<wall.size(); row++)
{
int edge_postion = 0;
for(int brick_no=0; brick_no< wall[row].size() -1; brick_no++)
{
int current_brick_length = wall[row][brick_no];
edge_postion = edge_postion + current_brick_length ;
edge_frequency[edge_postion]++;
max_frequency = max(edge_frequency[edge_postion],max_frequency);
}
}
return wall.size() - max_frequency;
}
};```Code language: PHP (php)```

## Brick Wall SolutionJava

``````class Solution {
public int leastBricks(List<List<Integer>> wall) {
var edge_frequency=new HashMap<Integer,Integer>();
int max_frequency = 0;
for(int row=0; row<wall.size(); row++)
{
int edge_postion = 0;
for(int brick_no=0; brick_no< wall.get(row).size() -1; brick_no++)
{
int current_brick_length = wall.get(row).get(brick_no);
edge_postion = edge_postion + current_brick_length ;
edge_frequency.put(edge_postion,edge_frequency.getOrDefault(edge_postion,0)+1);
max_frequency = Math.max(edge_frequency.get(edge_postion),max_frequency);
}
}
return wall.size() - max_frequency;
}
}```Code language: PHP (php)```

## Brick Wall SolutionJavaScript

``````var leastBricks = function(wall) {
let freq = new Map(), best = 0
for (let i = 0; i < wall.length; i++) {
let row = wall[i], rowSum = row[0]
for (let j = 1; j < row.length; j++) {
freq.set(rowSum, (freq.get(rowSum) || 0) + 1)
rowSum += row[j]
}
}
for (let [k,v] of freq)
if (v > best) best = v
return wall.length - best
};```Code language: JavaScript (javascript)```

## Brick Wall SolutionPython

``````class Solution(object):
def leastBricks(self, wall):
d = collections.defaultdict(int)
for line in wall:
i = 0
for brick in line[:-1]:
i += brick
d[i] += 1
# print len(wall), d
return len(wall)-max(d.values()+[0])``````
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