<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>

/*

1 3 5 6
3 4 6
1 4 6
2 6
3 4 6
1 4 5 6

*/

class Solution {
public:
    int leastBricks(vector<vector<int>>& wall) {
        int n = wall.size();
        map<int, int> hash;
        vector<int> freq;

        for(auto &w : wall){
            // accumulate the number
            for(int i=1; i<w.size(); i++){
                w[i] += w[i-1];
            }
            // record the frequency
            for(int i=0; i<w.size(); i++){
                hash[w[i]]++;
            }
        }

        if(hash.size() == 1) return n;

        for(auto &[a, b] : hash) freq.push_back(b);
        sort(freq.begin(), freq.end(), greater<int>());

        return n - freq[1];
    }
};
26800