Best meeting point in 2D binary array

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++ program to find best meeting point in 2D array*/
#include <bits/stdc++.h>
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 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<Integer> vertical = new ArrayList<>(); 
    List<Integer> 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

# 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# 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

<script>
/* Javascript program to find best 
meeting point in 2D array*/

let ROW = 3;
let COL =5 ;

function minTotalDistance(grid)
{
    if (ROW == 0 || COL == 0) 
        return 0; 
      
    let vertical = [];
    let horizontal = [];
      
    // Find all members home's position 
    for (let i = 0; i < ROW; i++)
    { 
        for (let j = 0; j < COL; j++) 
        { 
            if (grid[i][j] == 1) 
            { 
                vertical.push(i); 
                horizontal.push(j); 
            } 
        } 
    } 
      
    // Sort positions so we can find most 
    // beneficial point 
    (vertical).sort(function(a,b){return a-b;}); 
    (horizontal).sort(function(a,b){return a-b;}); 
      
    // middle position will always beneficial 
    // for all group members but it will be 
    // sorted which we have already done 
    let size = vertical.length / 2; 
    let x = vertical[size]; 
    let y = horizontal[size]; 
      
    // Now find total distance from best meeting 
    // point (x,y) using Manhattan Distance formula 
    let distance = 0; 
    for (let i = 0; i < ROW; i++) 
        for (let j = 0; j < COL; j++) 
            if (grid[i][j] == 1) 
                distance += Math.abs(x - i) +
                Math.abs(y - j); 
      
    return distance; 
}

// Driver code
let grid = [[1, 0, 1, 0, 1], 
                    [0, 1, 0, 0, 0],
                    [0, 1, 1, 0, 0]];
document.write(minTotalDistance(grid));


// This code is contributed by rag2127
</script>

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)