feat(work): add Word Ladder and Rotting Oranges solutions

work/1/Medium/Array/994.rotting-oranges.py

@@ -0,0 +1,93 @@
+"""
+994. Rotting Oranges
+Difficulty: Medium
+https://leetcode.com/problems/rotting-oranges/
+
+──────────────────────────────────────────────────
+
+You are given an m x n grid where each cell can have one of three
+values:
+
+        • 0 representing an empty cell,
+
+        • 1 representing a fresh orange, or
+
+        • 2 representing a rotten orange.
+
+Every minute, any fresh orange that is 4-directionally adjacent 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.
+
+
+
+Example 1:
+
+Input: grid = [[2,1,1],[1,1,0],[0,1,1]]
+Output: 4
+
+Example 2:
+
+Input: grid = [[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: grid = [[0,2]]
+Output: 0
+Explanation: Since there are already no fresh oranges at minute 0,
+the answer is just 0.
+
+
+
+Constraints:
+
+        • m == grid.length
+
+        • n == grid[i].length
+
+        • 1 <= m, n <= 10
+
+        • grid[i][j] is 0, 1, or 2.
+"""
+
+from collections import deque
+
+
+class Solution:
+    def orangesRotting(self, grid: list[list[int]]) -> int:
+        q = deque()
+        time, fresh = 0, 0
+
+        rows, cols = len(grid), len(grid[0])
+        for r in range(rows):
+            for c in range(cols):
+                if grid[r][c] == 1:
+                    fresh += 1
+                if grid[r][c] == 2:
+                    q.append([r, c])
+
+        directions = [[0, 1], [0, -1], [1, 0], [-1, 0]]
+
+        while q and fresh > 0:
+            for i in range(len(q)):
+                r, c = q.popleft()
+                for dr, dc in directions:
+                    row, col = dr + r, dc + c
+                    # check if in bounds
+                    if (
+                        row < 0
+                        or col < 0
+                        or row == len(grid)
+                        or col == len(grid[0])
+                        or grid[row][col] != 1
+                    ):
+                        continue
+                    grid[row][col] = 2
+                    q.append([row, col])
+                    fresh -= 1
+            time += 1
+        return time if fresh == 0 else -1