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‎Problemset/0079-Word Search/README.md

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# 79. Word Search
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Given an `m x n` grid of characters `board` and a string `word`, return `true` *if* `word` *exists in the grid*.
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The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
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#### Example 1:
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![](https://assets.leetcode.com/uploads/2020/11/04/word2.jpg)
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<pre>
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<strong>Input:</strong> board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
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<strong>Output:</strong> true
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</pre>
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#### Example 2:
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![](https://assets.leetcode.com/uploads/2020/11/04/word-1.jpg)
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<pre>
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<strong>Input:</strong> board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
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<strong>Output:</strong> true
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</pre>
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#### Example 3:
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![](https://assets.leetcode.com/uploads/2020/10/15/word3.jpg)
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<pre>
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<strong>Input:</strong> board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
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<strong>Output:</strong> false
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</pre>
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#### Constraints:
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* `m == board.length`
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* `n = board[i].length`
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* `1 <= m, n <= 6`
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* `1 <= word.length <= 15`
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* `board` and `word` consists of only lowercase and uppercase English letters.
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**Follow up:** Could you use search pruning to make your solution faster with a larger `board`?
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## Solutions (Python)
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### 1. Solution
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```Python
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class Solution:
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def exist(self, board: List[List[str]], word: str) -> bool:
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m, n = len(board), len(board[0])
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for r in range(m):
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for c in range(n):
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if board[r][c] != word[0]:
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continue
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visited = {(r, c)}
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stack = [[r, c, 0, 0]]
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while stack != []:
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i, j, k, l = stack[-1]
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if k == len(word) - 1:
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return True
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if l == 0:
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stack[-1][3] += 1
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if i > 0 and (i - 1, j) not in visited and board[i - 1][j] == word[k + 1]:
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visited.add((i - 1, j))
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stack.append([i - 1, j, k + 1, 0])
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elif l == 1:
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stack[-1][3] += 1
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if i < m - 1 and (i + 1, j) not in visited and board[i + 1][j] == word[k + 1]:
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visited.add((i + 1, j))
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stack.append([i + 1, j, k + 1, 0])
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elif l == 2:
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stack[-1][3] += 1
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if j > 0 and (i, j - 1) not in visited and board[i][j - 1] == word[k + 1]:
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visited.add((i, j - 1))
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stack.append([i, j - 1, k + 1, 0])
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elif l == 3:
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stack[-1][3] += 1
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if j < n - 1 and (i, j + 1) not in visited and board[i][j + 1] == word[k + 1]:
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visited.add((i, j + 1))
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stack.append([i, j + 1, k + 1, 0])
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else:
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stack.pop()
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visited.remove((i, j))
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return False
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```

‎Problemset/0079-Word Search/README_CN.md

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# 79. 单词搜索
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给定一个 `m x n` 二维字符网格 `board` 和一个字符串单词 `word` 。如果 `word` 存在于网格中，返回 `true` ；否则，返回 `false` 。
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单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
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#### 示例 1:
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![](https://assets.leetcode.com/uploads/2020/11/04/word2.jpg)
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<pre>
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<strong>输入:</strong> board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
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<strong>输出:</strong> true
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</pre>
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#### 示例 2:
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![](https://assets.leetcode.com/uploads/2020/11/04/word-1.jpg)
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<pre>
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<strong>输入:</strong> board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
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<strong>输出:</strong> true
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</pre>
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#### 示例 3:
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![](https://assets.leetcode.com/uploads/2020/10/15/word3.jpg)
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<pre>
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<strong>输入:</strong> board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
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<strong>输出:</strong> false
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</pre>
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#### 提示:
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* `m == board.length`
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* `n = board[i].length`
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* `1 <= m, n <= 6`
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* `1 <= word.length <= 15`
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* `board` 和 `word` 仅由大小写英文字母组成
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**进阶：**你可以使用搜索剪枝的技术来优化解决方案，使其在 `board` 更大的情况下可以更快解决问题？
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## 题解 (Python)
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### 1. 题解
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```Python
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class Solution:
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def exist(self, board: List[List[str]], word: str) -> bool:
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m, n = len(board), len(board[0])
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for r in range(m):
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for c in range(n):
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if board[r][c] != word[0]:
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continue
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visited = {(r, c)}
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stack = [[r, c, 0, 0]]
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while stack != []:
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i, j, k, l = stack[-1]
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if k == len(word) - 1:
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return True
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if l == 0:
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stack[-1][3] += 1
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if i > 0 and (i - 1, j) not in visited and board[i - 1][j] == word[k + 1]:
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visited.add((i - 1, j))
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stack.append([i - 1, j, k + 1, 0])
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elif l == 1:
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stack[-1][3] += 1
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if i < m - 1 and (i + 1, j) not in visited and board[i + 1][j] == word[k + 1]:
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visited.add((i + 1, j))
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stack.append([i + 1, j, k + 1, 0])
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elif l == 2:
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stack[-1][3] += 1
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if j > 0 and (i, j - 1) not in visited and board[i][j - 1] == word[k + 1]:
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visited.add((i, j - 1))
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stack.append([i, j - 1, k + 1, 0])
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elif l == 3:
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stack[-1][3] += 1
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if j < n - 1 and (i, j + 1) not in visited and board[i][j + 1] == word[k + 1]:
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visited.add((i, j + 1))
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stack.append([i, j + 1, k + 1, 0])
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else:
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stack.pop()
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visited.remove((i, j))
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return False
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```

‎Problemset/0079-Word Search/Solution.py

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class Solution:
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def exist(self, board: List[List[str]], word: str) -> bool:
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m, n = len(board), len(board[0])
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for r in range(m):
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for c in range(n):
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if board[r][c] != word[0]:
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continue
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visited = {(r, c)}
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stack = [[r, c, 0, 0]]
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while stack != []:
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i, j, k, l = stack[-1]
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if k == len(word) - 1:
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return True
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if l == 0:
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stack[-1][3] += 1
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if i > 0 and (i - 1, j) not in visited and board[i - 1][j] == word[k + 1]:
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visited.add((i - 1, j))
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stack.append([i - 1, j, k + 1, 0])
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elif l == 1:
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stack[-1][3] += 1
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if i < m - 1 and (i + 1, j) not in visited and board[i + 1][j] == word[k + 1]:
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visited.add((i + 1, j))
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stack.append([i + 1, j, k + 1, 0])
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elif l == 2:
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stack[-1][3] += 1
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if j > 0 and (i, j - 1) not in visited and board[i][j - 1] == word[k + 1]:
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visited.add((i, j - 1))
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stack.append([i, j - 1, k + 1, 0])
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elif l == 3:
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stack[-1][3] += 1
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if j < n - 1 and (i, j + 1) not in visited and board[i][j + 1] == word[k + 1]:
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visited.add((i, j + 1))
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stack.append([i, j + 1, k + 1, 0])
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else:
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stack.pop()
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visited.remove((i, j))
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return False

‎README.md

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[76][76l] |[Minimum Window Substring][76] |![rs]
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[77][77l] |[Combinations][77] |![py]
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[78][78l] |[Subsets][78] |![rs]
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[79][79l] |[Word Search][79] |![py]
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[80][80l] |[Remove Duplicates from Sorted Array II][80] |![rs]
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[81][81l] |[Search in Rotated Sorted Array II][81] |![rs]
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[82][82l] |[Remove Duplicates from Sorted List II][82] |![rb]
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[76]:Problemset/0076-Minimum%20Window%20Substring/README.md#76-minimum-window-substring
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[77]:Problemset/0077-Combinations/README.md#77-combinations
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[78]:Problemset/0078-Subsets/README.md#78-subsets
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[79]:Problemset/0079-Word%20Search/README.md#79-word-search
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[80]:Problemset/0080-Remove%20Duplicates%20from%20Sorted%20Array%20II/README.md#80-remove-duplicates-from-sorted-array-ii
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[81]:Problemset/0081-Search%20in%20Rotated%20Sorted%20Array%20II/README.md#81-search-in-rotated-sorted-array-ii
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[82]:Problemset/0082-Remove%20Duplicates%20from%20Sorted%20List%20II/README.md#82-remove-duplicates-from-sorted-list-ii
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[76l]:https://leetcode.com/problems/minimum-window-substring/
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[77l]:https://leetcode.com/problems/combinations/
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[78l]:https://leetcode.com/problems/subsets/
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[79l]:https://leetcode.com/problems/word-search/
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[80l]:https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/
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[81l]:https://leetcode.com/problems/search-in-rotated-sorted-array-ii/
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[82l]:https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/

‎README_CN.md

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[76][76l] |[最小覆盖子串][76] |![rs]
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[77][77l] |[组合][77] |![py]
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[78][78l] |[子集][78] |![rs]
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[79][79l] |[单词搜索][79] |![py]
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[80][80l] |[删除排序数组中的重复项 II][80] |![rs]
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[81][81l] |[搜索旋转排序数组 II][81] |![rs]
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[82][82l] |[删除排序链表中的重复元素 II][82] |![rb]
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[76]:Problemset/0076-Minimum%20Window%20Substring/README_CN.md#76-最小覆盖子串
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[77]:Problemset/0077-Combinations/README_CN.md#77-组合
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[78]:Problemset/0078-Subsets/README_CN.md#78-子集
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[79]:Problemset/0079-Word%20Search/README_CN.md#79-单词搜索
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[80]:Problemset/0080-Remove%20Duplicates%20from%20Sorted%20Array%20II/README_CN.md#80-删除排序数组中的重复项-ii
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[81]:Problemset/0081-Search%20in%20Rotated%20Sorted%20Array%20II/README_CN.md#81-搜索旋转排序数组-ii
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[82]:Problemset/0082-Remove%20Duplicates%20from%20Sorted%20List%20II/README_CN.md#82-删除排序链表中的重复元素-ii
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[76l]:https://leetcode.cn/problems/minimum-window-substring/
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[77l]:https://leetcode.cn/problems/combinations/
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[78l]:https://leetcode.cn/problems/subsets/
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[79l]:https://leetcode.cn/problems/word-search/
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[80l]:https://leetcode.cn/problems/remove-duplicates-from-sorted-array-ii/
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[81l]:https://leetcode.cn/problems/search-in-rotated-sorted-array-ii/
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[82l]:https://leetcode.cn/problems/remove-duplicates-from-sorted-list-ii/

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