<p>There is a rectangular brick wall in front of you with <code>n</code> rows of bricks. The <code>i<sup>th</sup></code> 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.</p>

<p>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.</p>

<p>Given the 2D array <code>wall</code> that contains the information about the wall, return <em>the minimum number of crossed bricks after drawing such a vertical line</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/cutwall-grid.jpg" style="width: 493px; height: 577px;" />
<pre>
<strong>Input:</strong> wall = [[1,2,2,1],[3,1,2],[1,3,2],[2,4],[3,1,2],[1,3,1,1]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> wall = [[1],[1],[1]]
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == wall.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= wall[i].length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= sum(wall[i].length) &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>sum(wall[i])</code> is the same for each row <code>i</code>.</li>
	<li><code>1 &lt;= wall[i][j] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution {
public:
    int leastBricks(vector<vector<int>>& w) 
    {
        if(w.size()==1) 
        {
            if(w[0].size()==1) return 1;
            return 0;
        }
        for(int i=0;i<w.size();i++)
        {
            for(int j=1;j<w[i].size();j++) w[i][j] = w[i][j]+w[i][j-1];
        }
        int mx =0;
        for(int i=0;i<w.size();i++)
        {
           for(int k=0;k<w[i].size()-1;k++)
           {
             int s = w[i][k];
            int t=0;
            for(int j=0;j<w.size();j++)
            {
               if(binary_search(w[j].begin(), w[j].end(), s)&&(w[j][w[j].size()-1]!=s)) t++;
            }

            mx = max(mx,t);
           }
        }
        return w.size() - mx;
    }
};
23100