# Best meeting point in 2D binary array

• Difficulty Level : Medium
• Last Updated : 18 Jul, 2022

You are given a 2D grid of values 0 or 1, where each 1 marks the home of someone in a group. And the group of two or more people wants to meet and minimize the total travel distance. They can meet anywhere means that there might be a home or not.

• The distance is calculated using Manhattan Distance, where distance(p1, p2) = |p2.x – p1.x| + |p2.y – p1.y|.
• Find the total distance that needs to be traveled to reach the best meeting point (Total distance traveled is minimum).

Examples:

```Input : grid[][] = {{1, 0, 0, 0, 1},
{0, 0, 0, 0, 0},
{0, 0, 1, 0, 0}};
Output : 6
Best meeting point is (0, 2).
Total distance traveled is 2 + 2 + 2 = 6

Input : grid[3][5] = {{1, 0, 1, 0, 1},
{0, 1, 0, 0, 0},
{0, 1, 1, 0, 0}};
Output : 11```

Concept :-

Let the group of people be present at the coordinates G : (Gx, Gy), H : (Hx, Hy), J : (Jx, Jy), and K : (Kx, Ky).

Let the Best Meeting Point be P : (x, y)

Distance of every point from P will be given by,

`D = |Gx - x| + |Gy - y| + |Hx - x| + |Hy - y| + |Jx - x| + |Jy - y| + |Kx - x| + |Ky - y|`

which can also be written as,

```D = Dx + Dy
where, Dx  =  |Gx - x|  + |Hx - x| + |Jx - x|  + |Kx - x|
and, Dy = |Gy - y| + |Hy - y| + |Jy - y| + |Ky - y|```

To minimize D, we should minimize Dx and Dy.

```Dx will be minimum if x is the median of (Gx, Hx, Jx, Kx)
and Similarly, Dy will be minimum if y is the median of (Gy, Hy, Jy, Ky)```

Steps :-

1. Store all horizontal and vertical positions of all group member.
2. Now sort it to find minimum middle position, which will be the best meeting point.
3. Find the distance of all members from best meeting point.

For example in above diagram, horizontal positions are {0, 2, 0} and vertical positions are {0, 2, 4}. After sorting both, we get {0, 0, 2} and {0, 2, 4}. Middle point is (0, 2).

Note : Even no. of 1’s have two middle points, then also it works. Two middle points means it have two best meeting points always. Both cases will give same distance. So we will consider only one best meeting point to avoid the more overhead, Because our aim is to find the distance only.

Implementation:

## C++

 `/* C++ program to find best meeting point in 2D array*/``#include ``using` `namespace` `std;``#define ROW 3``#define COL 5` `int` `minTotalDistance(``int` `grid[][COL]) {``    ``if` `(ROW == 0 || COL == 0)``         ``return` `0;``    ` `    ``vector<``int``> vertical;``    ``vector<``int``> horizontal;``    ` `    ``// Find all members home's position``    ``for` `(``int` `i = 0; i < ROW; i++) {``        ``for` `(``int` `j = 0; j < COL; j++) {``            ``if` `(grid[i][j] == 1) {``                ``vertical.push_back(i);``                ``horizontal.push_back(j);``            ``}``        ``}``    ``}``    ` `    ``// Sort positions so we can find most``    ``// beneficial point``    ``sort(vertical.begin(),vertical.end());``    ``sort(horizontal.begin(),horizontal.end());``    ` `    ``// middle position will always beneficial``    ``// for all group members but it will be``    ``// sorted which we have already done``    ``int` `size = vertical.size()/2;``    ``int` `x = vertical[size];``    ``int` `y = horizontal[size];``    ` `    ``// Now find total distance from best meeting``    ``// point (x,y) using Manhattan Distance formula``    ``int` `distance = 0;``    ``for` `(``int` `i = 0; i < ROW; i++)``        ``for` `(``int` `j = 0; j < COL; j++)``            ``if` `(grid[i][j] == 1)``                ``distance += ``abs``(x - i) + ``abs``(y - j);``    ` `    ``return` `distance;``}` `// Driver program to test above functions``int` `main() {``    ``int` `grid[ROW][COL] = {{1, 0, 1, 0, 1}, {0, 1, 0, 0, 0},{0, 1, 1, 0, 0}};``    ``cout << minTotalDistance(grid);``    ``return` `0;``}`

## Java

 `/* Java program to find best``meeting point in 2D array*/``import` `java.util.*;` `class` `GFG``{` `static` `int` `minTotalDistance(``int` `grid[][],``int` `ROW , ``int` `COL)``{``    ``if` `(ROW == ``0` `|| COL == ``0``)``        ``return` `0``;``    ` `    ``List vertical = ``new` `ArrayList<>();``    ``List horizontal = ``new` `ArrayList<>();``    ` `    ``// Find all members home's position``    ``for` `(``int` `i = ``0``; i < ROW; i++)``    ``{``        ``for` `(``int` `j = ``0``; j < COL; j++)``        ``{``            ``if` `(grid[i][j] == ``1``)``            ``{``                ``vertical.add(i);``                ``horizontal.add(j);``            ``}``        ``}``    ``}``    ` `    ``// Sort positions so we can find most``    ``// beneficial point``    ``Collections.sort(vertical);``    ``Collections.sort(horizontal);``    ` `    ``// middle position will always beneficial``    ``// for all group members but it will be``    ``// sorted which we have already done``    ``int` `size = vertical.size() / ``2``;``    ``int` `x = vertical.get(size);``    ``int` `y = horizontal.get(size);``    ` `    ``// Now find total distance from best meeting``    ``// point (x,y) using Manhattan Distance formula``    ``int` `distance = ``0``;``    ``for` `(``int` `i = ``0``; i < ROW; i++)``        ``for` `(``int` `j = ``0``; j < COL; j++)``            ``if` `(grid[i][j] == ``1``)``                ``distance += Math.abs(x - i) +``                ``Math.abs(y - j);``    ` `    ``return` `distance;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `grid[][] = {{``1``, ``0``, ``1``, ``0``, ``1``},``                    ``{``0``, ``1``, ``0``, ``0``, ``0``},``                    ``{``0``, ``1``, ``1``, ``0``, ``0``}};``    ``System.out.println(minTotalDistance(grid , grid.length , grid[``0``].length));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python program to find best meeting point in 2D array``ROW ``=` `3``COL ``=` `5` `def` `minTotalDistance(grid: ``list``) ``-``> ``int``:``    ``if` `ROW ``=``=` `0` `or` `COL ``=``=` `0``:``        ``return` `0` `    ``vertical ``=` `[]``    ``horizontal ``=` `[]` `    ``# Find all members home's position``    ``for` `i ``in` `range``(ROW):``        ``for` `j ``in` `range``(COL):``            ``if` `grid[i][j] ``=``=` `1``:``                ``vertical.append(i)``                ``horizontal.append(j)` `    ``# Sort positions so we can find most``    ``# beneficial point``    ``vertical.sort()``    ``horizontal.sort()` `    ``# middle position will always beneficial``    ``# for all group members but it will be``    ``# sorted which we have already done``    ``size ``=` `len``(vertical) ``/``/` `2``    ``x ``=` `vertical[size]``    ``y ``=` `horizontal[size]` `    ``# Now find total distance from best meeting``    ``# point (x,y) using Manhattan Distance formula``    ``distance ``=` `0``    ``for` `i ``in` `range``(ROW):``        ``for` `j ``in` `range``(COL):``            ``if` `grid[i][j] ``=``=` `1``:``                ``distance ``+``=` `abs``(x ``-` `i) ``+` `abs``(y ``-` `j)` `    ``return` `distance` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``grid ``=` `[[``1``, ``0``, ``1``, ``0``, ``1``],``            ``[``0``, ``1``, ``0``, ``0``, ``0``],``            ``[``0``, ``1``, ``1``, ``0``, ``0``]]``    ``print``(minTotalDistance(grid))` `# This code is contributed by``# sanjeev2552`

## C#

 `/* C# program to find best``meeting point in 2D array*/``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ` `static` `int` `ROW = 3;``static` `int` `COL = 5 ;` `static` `int` `minTotalDistance(``int` `[,]grid)``{``    ``if` `(ROW == 0 || COL == 0)``        ``return` `0;``    ` `    ``List<``int``> vertical = ``new` `List<``int``>();``    ``List<``int``> horizontal = ``new` `List<``int``>();``    ` `    ``// Find all members home's position``    ``for` `(``int` `i = 0; i < ROW; i++)``    ``{``        ``for` `(``int` `j = 0; j < COL; j++)``        ``{``            ``if` `(grid[i, j] == 1)``            ``{``                ``vertical.Add(i);``                ``horizontal.Add(j);``            ``}``        ``}``    ``}``    ` `    ``// Sort positions so we can find most``    ``// beneficial point``    ``vertical.Sort();``    ``horizontal.Sort();``    ` `    ``// middle position will always beneficial``    ``// for all group members but it will be``    ``// sorted which we have already done``    ``int` `size = vertical.Count / 2;``    ``int` `x = vertical[size];``    ``int` `y = horizontal[size];``    ` `    ``// Now find total distance from best meeting``    ``// point (x,y) using Manhattan Distance formula``    ``int` `distance = 0;``    ``for` `(``int` `i = 0; i < ROW; i++)``        ``for` `(``int` `j = 0; j < COL; j++)``            ``if` `(grid[i, j] == 1)``                ``distance += Math.Abs(x - i) +``                ``Math.Abs(y - j);``    ` `    ``return` `distance;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[,]grid = {{1, 0, 1, 0, 1},``                    ``{0, 1, 0, 0, 0},``                    ``{0, 1, 1, 0, 0}};``    ``Console.WriteLine(minTotalDistance(grid));``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output

`11`

In worst case if all the elements of Matrix are set to 1. Then we would have to sort N*M elements. Time to sort is (N*M)log(N*M) or 2N^2log(N)

Time Complexity : O(M*N)
Auxiliary Space : O(N)

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