994 Rotting Oranges.py

Author: rajeevranjancom

994 Rotting Oranges.py

@@ -0,0 +1,73 @@
+#!/usr/bin/python3
+"""
+In a given grid, each cell can have one of three values:
+
+the value 0 representing an empty cell;
+the value 1 representing a fresh orange;
+the value 2 representing a rotten orange.
+Every minute, any fresh orange that is adjacent (4-directionally) to a rotten
+orange becomes rotten.
+
+Return the minimum number of minutes that must elapse until no cell has a fresh
+orange.  If this is impossible, return -1 instead.
+
+Example 1:
+
+Input: [[2,1,1],[1,1,0],[0,1,1]]
+Output: 4
+Example 2:
+
+Input: [[2,1,1],[0,1,1],[1,0,1]]
+Output: -1
+Explanation:  The orange in the bottom left corner (row 2, column 0) is never
+rotten, because rotting only happens 4-directionally.
+Example 3:
+
+Input: [[0,2]]
+Output: 0
+Explanation:  Since there are already no fresh oranges at minute 0, the answer
+is just 0.
+
+Note:
+1 <= grid.length <= 10
+1 <= grid[0].length <= 10
+grid[i][j] is only 0, 1, or 2.
+"""
+from typing import List
+
+
+dirs = ((0, -1), (0, 1), (-1, 0), (1, 0))
+
+
+class Solution:
+    def orangesRotting(self, grid: List[List[int]]) -> int:
+        """
+        maintain a q for the newly rotten
+        """
+        m, n = len(grid), len(grid[0])
+        q = []
+        for i in range(m):
+            for j in range(n):
+                if grid[i][j] == 2:
+                    q.append((i, j))
+
+        t = -1
+        while q:
+            t += 1
+            cur_q = []
+            for i, j in q:
+                for di, dj in dirs:
+                    I = i + di
+                    J = j + dj
+                    if 0 <= I < m and 0 <= J < n and grid[I][J] == 1:
+                        grid[I][J] = 2
+                        cur_q.append((I, J))
+            q = cur_q
+
+        has_fresh = any(
+            grid[i][j] == 1
+            for i in range(m)
+            for j in range(n)
+        )
+
+        return max(0, t) if not has_fresh else -1