# [leetcode] Best Meeting Point

Best Meeting Point

A group of people wants to meet and minimize the total travel distance. You are given a 2D grid of values 0 or 1, where each 1 marks the home of someone in the group. The distance is calculated using Manhattan Distance, where distance(p1, p2) = `|p2.x - p1.x| + |p2.y - p1.y|`.

For example, given three people living at `(0,0)`, `(0,4)`, and `(2,2)`:

```1 - 0 - 0 - 0 - 1
|   |   |   |   |
0 - 0 - 0 - 0 - 0
|   |   |   |   |
0 - 0 - 1 - 0 - 0```

The point `(0,2)` is an ideal meeting point, as the total travel distance of 2+2+2=6 is minimal. So return 6.

Dynamic programming.

Brute force solution is trivial. O(n^4).

This solution optimize it to O(n^2)

Decompose the 2D manhattan distance into two 1D manhattan distance, since the two dimension is individual.

```class Solution {
public:
int INTMAX = 0x7fffffff;
int minTotalDistance(vector<vector<int>>& grid) {
if(grid.size() == 0 || grid.size() == 0) return 0;
int rows = grid.size();
int columns = grid.size();
int distance;
int countC[columns]; // number of 1s in each column
int countR[rows]; // number of 1s in each row

int sumRestR[rows]; // sumRestR[i] = all the
int sumRestC[columns];

for(int r = 0; r < rows; r++){
countR[r] = 0;
for(int c = 0; c < columns; c++){
if(grid[r][c] == 1) countR[r]++;
}
}

for(int c = 0; c < columns; c++){
countC[c] = 0;
for(int r = 0; r < rows; r++){
if(grid[r][c] == 1) countC[c]++;
}
}

for(int i = 0; i < rows; i++){
sumRestR[i] = 0;
for(int j = 0; j < rows; j++){
sumRestR[i] += abs(j - i) * countR[j];
}
}

for(int i = 0; i < columns; i++){
sumRestC[i] = 0;
for(int j = 0; j < columns; j++){
sumRestC[i] += abs(j - i) * countC[j];
}
}
distance = INTMAX;
for(int r = 0; r < rows; r++){
for(int c = 0; c < columns; c++){
distance = min(distance, sumRestR[r] + sumRestC[c]);
}
}
return distance;
}
};```

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