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

work/1/Hard/Hash Table/127.word-ladder.py

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+"""
+127. Word Ladder
+Difficulty: Hard
+https://leetcode.com/problems/word-ladder/
+
+──────────────────────────────────────────────────
+
+A transformation sequence from word beginWord to word endWord using a
+dictionary wordList is a sequence of words beginWord -> s1 -> s2 ->
+... -> sk such that:
+
+        • Every adjacent pair of words differs by a single letter.
+
+• Every si for 1 <= i <= k is in wordList. Note that beginWord does
+not need to be in wordList.
+
+        • sk == endWord
+
+Given two words, beginWord and endWord, and a dictionary wordList,
+return the number of words in the shortest transformation sequence
+from beginWord to endWord, or 0 if no such sequence exists.
+
+
+
+Example 1:
+
+Input: beginWord = "hit", endWord = "cog", wordList =
+["hot","dot","dog","lot","log","cog"]
+Output: 5
+Explanation: One shortest transformation sequence is "hit" -> "hot"
+-> "dot" -> "dog" -> cog", which is 5 words long.
+
+Example 2:
+
+Input: beginWord = "hit", endWord = "cog", wordList =
+["hot","dot","dog","lot","log"]
+Output: 0
+Explanation: The endWord "cog" is not in wordList, therefore there is
+no valid transformation sequence.
+
+
+
+Constraints:
+
+        • 1 <= beginWord.length <= 10
+
+        • endWord.length == beginWord.length
+
+        • 1 <= wordList.length <= 5000
+
+        • wordList[i].length == beginWord.length
+
+• beginWord, endWord, and wordList[i] consist of lowercase English
+letters.
+
+        • beginWord != endWord
+
+        • All the words in wordList are unique.
+"""
+
+import collections
+
+
+class Solution:
+    def ladderLength(self, beginWord: str, endWord: str, wordList: list[str]) -> int:
+        if endWord not in wordList:
+            return 0
+
+        adj = collections.defaultdict(list)
+        wordList.append(beginWord)
+
+        for word in wordList:
+            for j in range(len(word)):
+                pattern = word[:j] + "*" + word[j + 1 :]
+                adj[pattern].append(word)
+
+        visit = set([beginWord])
+        q = collections.deque([beginWord])
+        res = 1
+
+        while q:
+            for i in range(len(q)):
+                word = q.popleft()
+                if word == endWord:
+                    return res
+                for j in range(len(word)):
+                    pattern = word[:j] + "*" + word[j + 1 :]
+                    for adjWord in adj[pattern]:
+                        if adjWord not in visit:
+                            visit.add(adjWord)
+                            q.append(adjWord)
+            res += 1
+        return 0

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

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+"""
+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