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MatrixCellsinDistanceOrder.java
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/* (C) 2024 YourCompanyName */
package array;
import java.util.*;
/**
* Created by gouthamvidyapradhan on 15/08/2019 We are given a matrix with R rows and C columns has
* cells with integer coordinates (r, c), where 0 <= r < R and 0 <= c < C.
*
* <p>Additionally, we are given a cell in that matrix with coordinates (r0, c0).
*
* <p>Return the coordinates of all cells in the matrix, sorted by their distance from (r0, c0) from
* smallest distance to largest distance. Here, the distance between two cells (r1, c1) and (r2, c2)
* is the Manhattan distance, |r1 - r2| + |c1 - c2|. (You may return the answer in any order that
* satisfies this condition.)
*
* <p>Example 1:
*
* <p>Input: R = 1, C = 2, r0 = 0, c0 = 0 Output: [[0,0],[0,1]] Explanation: The distances from (r0,
* c0) to other cells are: [0,1] Example 2:
*
* <p>Input: R = 2, C = 2, r0 = 0, c0 = 1 Output: [[0,1],[0,0],[1,1],[1,0]] Explanation: The
* distances from (r0, c0) to other cells are: [0,1,1,2] The answer [[0,1],[1,1],[0,0],[1,0]] would
* also be accepted as correct. Example 3:
*
* <p>Input: R = 2, C = 3, r0 = 1, c0 = 2 Output: [[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]] Explanation:
* The distances from (r0, c0) to other cells are: [0,1,1,2,2,3] There are other answers that would
* also be accepted as correct, such as [[1,2],[1,1],[0,2],[1,0],[0,1],[0,0]].
*
* <p>Note:
*
* <p>1 <= R <= 100 1 <= C <= 100 0 <= r0 < R 0 <= c0 < C
*
* <p>Solution: O (log (R x C)) Straight forward solution, find the Manhattan distance from the
* given cell to all the cells in the grid and sort by min distance and return their coordinates.
*/
public class MatrixCellsinDistanceOrder {
public static void main(String[] args) {
//
}
class Cell {
int max, i, j;
Cell(int max, int i, int j) {
this.max = max;
this.i = i;
this.j = j;
}
}
public int[][] allCellsDistOrder(int R, int C, int r0, int c0) {
List<Cell> list = new ArrayList<>();
for (int i = 0; i < R; i++) {
for (int j = 0; j < C; j++) {
int sum = Math.abs(r0 - i) + Math.abs(c0 - j);
list.add(new Cell(sum, i, j));
}
}
list.sort(Comparator.comparingInt(o -> o.max));
int[][] A = new int[list.size()][2];
for (int i = 0; i < A.length; i++) {
A[i][0] = list.get(i).i;
A[i][1] = list.get(i).j;
}
return A;
}
}