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Word Search
Difficulty: Medium
Category: DSA
Topics: Array, String, Backtracking, Depth-First Search, Matrix
Asked at: Amazon, Bloomberg, Intuit, Microsoft, Facebook, Snapchat, Apple, Oracle, eBay, Pinterest, Expedia
Given an `m x n` grid of characters `board` and a string `word`, return `true` _if_ `word` _exists in the grid_.
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.
**Example 1:**
**Input:** board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
**Output:** true
```
**Example 2:**
**Input:** board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
**Output:** true
```
**Example 3:**
**Input:** board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
**Output:** false
```
**Constraints:**
- `m == board.length`
- `n = board[i].length`
- `1 <= m, n <= 6`
- `1 <= word.length <= 15`
- `board` and `word` consists of only lowercase and uppercase English letters.
**Follow up:** Could you use search pruning to make your solution faster with a larger `board`?
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