<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>

 /*
    Approach: 
    Store the count of the end of the brick for each row in a hash and keep the track
    of max number of brick that ends at same position, return rows - max.
    
    Time complexity : O(n x m)
    Space complexity: O(n x m)

    n is number of rows, m is max brick in a row.
*/

class Solution {
public:
    int leastBricks(vector<vector<int>>& wall) {

        map<int,int> end_count;
        int end_of_brick, max_end_count=0;

        int rows = wall.size(),cols;

        for(int i =0;i<rows;i++){
            end_of_brick = 0;

            // '-1' because edge of the wall is not considered
            cols = wall[i].size() -1;
            for(int j =0;j<cols;j++){
                end_of_brick += wall[i][j];

                if(end_count.find(end_of_brick)!=end_count.end())
                {
                    end_count[end_of_brick]++;
                }
                else{
                    end_count[end_of_brick] = 1;
                }
                max_end_count = max(max_end_count,end_count[end_of_brick]);
            }
        }

        return rows - max_end_count;
    }
};
 
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