update 1030.matrix-cells-in-distance-order.java

Author: ketankr9

1030.matrix-cells-in-distance-order.java

@@ -0,0 +1,112 @@
+/*
+ * @lc app=leetcode id=1030 lang=java
+ *
+ * [1030] Matrix Cells in Distance Order
+ *
+ * https://leetcode.com/problems/matrix-cells-in-distance-order/description/
+ *
+ * algorithms
+ * Easy (65.08%)
+ * Likes:    173
+ * Dislikes: 83
+ * Total Accepted:    20.1K
+ * Total Submissions: 30.9K
+ * Testcase Example:  '1\n2\n0\n0'
+ *
+ * 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.
+ * 
+ * Additionally, we are given a cell in that matrix with coordinates (r0, c0).
+ * 
+ * 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.)
+ * 
+ * 
+ * 
+ * 
+ * Example 1:
+ * 
+ * 
+ * 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:
+ * 
+ * 
+ * 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:
+ * 
+ * 
+ * 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]].
+ * 
+ * 
+ * 
+ * 
+ * Note:
+ * 
+ * 
+ * 1 <= R <= 100
+ * 1 <= C <= 100
+ * 0 <= r0 < R
+ * 0 <= c0 < C
+ * 
+ * 
+ * 
+ * 
+ * 
+ */
+
+// @lc code=start
+class Solution {
+    public int[][] allCellsDistOrder(int R, int C, int r0, int c0) {
+        int[][] ans = new int[R*C][2];
+
+        int k = 1;
+        ans[0][0] = r0;
+        ans[0][1] = c0;
+        
+        for(int d=1; d<R+C; d++)
+        	for(int i=0; i<R; i++)
+        		for(int j=0; j<C; j++)
+        			if(Math.abs(i-r0) + Math.abs(j-c0) == d){
+        				ans[k][0] = i;
+        				ans[k][1] = j;
+        				k++;
+        			}
+        		
+        return ans;
+    }
+}
+        // int n = R*C, k = 0, d = 1;
+        // ans[k++] = {r0, c0};
+        // while(d < R+C-1){
+        // 	//up
+        // 	ans[k++] = {r0-d, c0};
+        // 	for(int y = c0+1, x = r0-d+1; y<c0+d && x<r0; y++, x++)
+        // 		ans[k++] = {x, y};
+        // 	//right
+        // 	for(int y = c0+d-1, x = r0+1; y>c0 && x<r0+d; y--, x++)
+        // 		ans[k++] = {x, y};
+        // 	//down
+        // 	for(int y = c0+1, x = r0-d+1; y<c0+d && x<r0; y--, x--)
+        // 		ans[k++] = {x, y};
+        // 	for(int y = c0+1, x = r0-d+1; y<c0+d && x<r0; y++, x++)
+        // 		ans[k++] = {x, y};
+        // }
+// @lc code=end