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BestMeetingPoint.java
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BestMeetingPoint.java
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/* (C) 2024 YourCompanyName */
package array;
/**
* Created by gouthamvidyapradhan on 10/03/2019 A group of two or more 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|.
*
* <p>Example:
*
* <p>Input:
*
* <p>1 - 0 - 0 - 0 - 1 | | | | | 0 - 0 - 0 - 0 - 0 | | | | | 0 - 0 - 1 - 0 - 0
*
* <p>Output: 6
*
* <p>Explanation: Given three people living at (0,0), (0,4), and (2,2): The point (0,2) is an ideal
* meeting point, as the total travel distance of 2+2+2=6 is minimal. So return 6.
*
* <p>Solution: O(N ^ 2 + M ^ 2) + O(N x M): Calculate the total number of persons in each row and
* each column and then take a minimum of cartesian product of each row and each column.
*/
public class BestMeetingPoint {
/**
* Main method
*
* @param args
*/
public static void main(String[] args) {
int[][] grid = {{1, 0, 0, 0, 1}, {0, 0, 0, 0, 0}, {0, 0, 1, 0, 0}};
System.out.println(new BestMeetingPoint().minTotalDistance(grid));
}
public int minTotalDistance(int[][] grid) {
int[] countR = new int[grid.length];
int[] countC = new int[grid[0].length];
int[] distR = new int[grid.length];
int[] distC = new int[grid[0].length];
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == 1) {
countR[i]++;
countC[j]++;
}
}
}
for (int i = 0; i < distR.length; i++) {
for (int j = 0; j < distR.length; j++) {
if (countR[j] != 0) {
distR[i] += Math.abs(j - i) * countR[j];
}
}
}
for (int i = 0; i < distC.length; i++) {
for (int j = 0; j < distC.length; j++) {
if (countC[j] != 0) {
distC[i] += Math.abs(j - i) * countC[j];
}
}
}
int min = Integer.MAX_VALUE;
for (int i = 0; i < distR.length; i++) {
for (int j = 0; j < distC.length; j++) {
min = Math.min(min, distR[i] + distC[j]);
}
}
return min;
}
}