calculate the brick's x coordinate and put in a map. and then get the maximum count and subtract from the layers.

```
class Solution {
public:
int leastBricks(vector<vector<int>>& wall) {
//add each layer to get the x position and count the max one!
unordered_map<int,int> mp;
for(int i=0;i<wall.size();i++)
{
int x=0;
for(int j=0;j<wall[i].size()-1;j++) //note the last one cannot be accounted
{
x+=wall[i][j];
mp[x]++;
}
}
//find the max
int maxlen=0;
for(auto it=mp.begin();it!=mp.end();it++) if(maxlen<it->second) maxlen=it->second;
return wall.size()-maxlen;
}
};
```