1030 Matrix Cells in Distance Order.py

Author: rajeevranjancom

1030 Matrix Cells in Distance Order.py

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+#!/usr/bin/python3
+"""
+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
+"""
+from typing import List
+
+
+class Solution:
+    def allCellsDistOrder(self, R: int, C: int, r0: int, c0: int) -> List[List[int]]:
+        """
+        bucket sort
+        """
+        r_max = max(r0, R-1 - r0)
+        c_max = max(c0, C-1 - c0)
+        lst = [[] for _ in range(r_max + c_max + 1)]
+        for i in range(R):
+            for j in range(C):
+                lst[abs(i - r0) + abs(j - c0)].append([i, j])
+
+        ret = []
+        for e in lst:
+            ret.extend(e)
+
+        return ret