# Minimum/maximum distance

The solution is from [1]. [1] also introduces several solutions for the variations of this problem.

`class Solution {public:    int minTotalDistance(vector<vector<int>>& grid) {        vector<int> x, y;        const int R = grid.size(), C = grid[0].size();        for(int i = 0; i<R;++i )            for(int j=0;j<C;++j)            {                if(grid[i][j]==1)                {                    x.push_back(j);                    y.push_back(i);                }            }        sort(x.begin(), x.end());        sort(y.begin(), y.end());        int mx = x[x.size()/2], my = y[y.size()/2];        int ret = 0;        for(int i = 0;i<x.size();++i)        {            ret += (abs(mx-x[i])+abs(my-y[i]));        }        return ret;    }};`

## 1515. Best Position for a Service Centre

See my previous post HERE.

# Reference

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## More from Jimmy Shen

Data Scientist/MLE/SWE @takemobi

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## Jimmy Shen

Data Scientist/MLE/SWE @takemobi