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

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.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

/* 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 alredy 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

/* Java program to find best 
meeting point in 2D array*/
import java.util.*;
  
class GFG 
{
      
static int ROW = 3;
static int COL =5 ;
  
static int minTotalDistance(int grid[][])
    if (ROW == 0 || COL == 0
        return 0
      
    Vector<Integer> vertical = new Vector<Integer>(); 
    Vector<Integer> horizontal = new Vector<Integer>();
      
    // 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 alredy 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));
}
  
// This code is contributed by 29AjayKumar

chevron_right


Output:

11

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

This article is contributed by Harshit Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : 29AjayKumar



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.