554. brick wall

You have a rectangle in front of you 、 from n A brick wall made of rows of bricks . These bricks are the same height （ That's one unit high ） But the width is different . The sum of the widths of each row of bricks is equal .

You're going to draw one now The top-down Of 、 Through the least Vertical line of brick . If you draw a line that just goes through the edge of the brick , Not through this brick . You can't draw a line along one of the two vertical edges of the wall , It's obviously not through a brick .

Give you a two-dimensional array wall , This array contains information about the wall . among ,wall[i] Is an array representing the width of each brick from left to right . You need to figure out how to draw this line The least number of bricks passed through , And back to The number of bricks passed through .

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

Tips ：

n == wall.length
1 <= n <= 104
1 <= wall[i].length <= 104
1 <= sum(wall[i].length) <= 2 * 104
 For each line  i ,sum(wall[i])  It's the same 
1 <= wall[i][j] <= 231 - 1

class Solution {
    
public:
    int leastBricks(vector<vector<int>>& wall) {
    
        map<int,int>m;

        int len=wall.size();
        for(int i=0;i<len;i++){
    
            int len2=wall[i].size();
            int x=0;
            for(int j=0;j<len2-1;j++){
    
                x+=wall[i][j];
                m[x]++;
            }
        }

        int max=0;

        for(map<int,int>::iterator it=m.begin();it!=m.end();it++){
    
            if(it->second>max)max=it->second;

        }
        return len-max;

    }
};