@@ -40,7 +40,8 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Image Smoother](problems/src/array/ImageSmoother.java) (Easy)
 - [Minimum Index Sum of Two Lists](problems/src/array/MinimumIndexSumOfTwoLists.java) (Easy)
 - [Card Flipping Game](problems/src/array/CardFilipGame.java) (Medium)
-- [Card Flipping Game](problems/src/array/EmployeeFreeTime.java) (Hard)
+- [Employee Free Time](problems/src/array/EmployeeFreeTime.java) (Hard)
+- [Best Meeting Point](problems/src/array/BestMeetingPoint.java) (Hard)
 
 #### [Backtracking](problems/src/backtracking)
 
@@ -181,7 +182,7 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Knight Probability in Chessboard](problems/src/dynamic_programming/KnightProbabilityInChessboard.java) (Medium)
 - [Largest Sum of Averages](problems/src/dynamic_programming/LargestSumOfAverages.java) (Medium)
 - [Minimum Number of Refueling Stops](problems/src/dynamic_programming/MinimumNumberOfRefuelingStops.java) (Hard)
-- [Minimum Number of Refueling Stops](problems/src/dynamic_programming/CatAndMouse.java) (Hard)
+- [Cat and Mouse](problems/src/dynamic_programming/CatAndMouse.java) (Hard)
 
 
 #### [Greedy](problems/src/greedy)

@@ -0,0 +1,81 @@
+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;
+}
+}