Word Search

Problem

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:

复制代码
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
Output: true

Example 2:

复制代码
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: true

Example 3:

复制代码
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: false

Intuition

The problem involves determining whether a given word exists in the provided grid. The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. Additionally, the same letter cell may not be used more than once. This problem can be solved using a depth-first search (DFS) approach.

Approach

DFS Function:

Implement a DFS function (dfs) that takes three parameters: the current row r, the current column c, and the index i representing the position in the word.

In the base case:

If i is equal to the length of the word, return True (the word has been found).

If the current cell (r, c) is out of bounds or has already been visited or contains a letter different from the corresponding letter in the word, return False.

Mark the current cell as visited by adding (r, c) to the set path.

Recursively call the dfs function for adjacent cells in all four directions (up, down, left, right) with the updated index i.

After the recursive calls, remove (r, c) from the set path to backtrack.

Iterate Over Grid:

Iterate over each cell in the grid and call the dfs function for each cell with the starting index i set to 0.

If the dfs function returns True for any cell, the word exists in the grid, and the function can return True.

Return Result:

If no cell results in a True return from the dfs function, return False, indicating that the word does not exist in the grid.

Complexity

  • Time complexity:

The time complexity is O(M * N * 4^L), where M is the number of rows, N is the number of columns, and L is the length of the word. The factor of 4^L accounts for the branching factor of the DFS.

  • Space complexity:

The space complexity is O(L), where L is the length of the word. This is due to the space required for the recursion stack and the set path used to track visited cells.

Code

复制代码
class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        rows, cols = len(board), len(board[0])
        path = set()

        def dfs(r, c, i):
            if i == len(word):
                return True
            if (r < 0 or c < 0 or
                r >= rows or c >= cols or
                word[i] != board[r][c] or (r, c) in path):
                return False

            path.add((r, c))
            res = (dfs(r + 1, c, i + 1) or
                    dfs(r - 1, c, i + 1) or
                    dfs(r, c - 1, i + 1) or
                    dfs(r, c + 1, i + 1))
                
            path.remove((r, c))
            return res

        for r in range(rows):
            for c in range(cols):
                if dfs(r, c, 0): return True
        return False
相关推荐
Miraitowa_cheems3 小时前
LeetCode算法日记 - Day 88: 环绕字符串中唯一的子字符串
java·数据结构·算法·leetcode·深度优先·动态规划
B站_计算机毕业设计之家3 小时前
python电商商品评论数据分析可视化系统 爬虫 数据采集 Flask框架 NLP情感分析 LDA主题分析 Bayes评论分类(源码) ✅
大数据·hadoop·爬虫·python·算法·数据分析·1024程序员节
小白菜又菜4 小时前
Leetcode 1518. Water Bottles
算法·leetcode·职场和发展
长存祈月心4 小时前
Rust Option 与 Result深度解析
算法
杭州杭州杭州5 小时前
机器学习(3)---线性算法,决策树,神经网络,支持向量机
算法·决策树·机器学习
不语n6 小时前
快速排序(Quick Sort)详解与图解
数据结构·算法·排序算法·快速排序·双指针排序
电鱼智能的电小鱼6 小时前
基于电鱼 ARM 工控机的AI视频智能分析方案:让传统监控变得更聪明
网络·arm开发·人工智能·嵌入式硬件·算法·音视频
初学者,亦行者7 小时前
Rust性能优化:内存对齐与缓存友好实战
算法·rust
py有趣7 小时前
LeetCode算法学习之杨辉三角
学习·算法·leetcode