Word Search - Solution & Explanation

One effective method to solve this problem is to use Depth-First Search (DFS) combined with backtracking. This approach involves starting from each cell and attempting to find the word by traversing adjacent cells, backtracking whenever a dead end is encountered.

DFS coupled with backtracking will be able to explore each possible path in the grid, returning true if a path spelling out the word is found, and false if all paths have been exhausted without success.

The function should mark a cell as visited before exploring further down its path and should unmark it (backtrack) when returning.

This solution uses a helper function, dfs(), to check character matches starting from each cell. The board cell is marked as visited by setting it to a null character. The function first checks boundary conditions and whether the target character matches the current cell before exploring in four directions (up, down, left, right). If the word is found, the function immediately returns true; otherwise, it backtracks.

An alternative but less optimal approach involves using Breadth-First Search (BFS). This approach can be useful for understanding the search pattern but is generally slower and less ideal compared to DFS for specific pathfinding problems.

Although BFS will attempt to explore each possibility layer-by-layer, it is less effective due to its nature of handling adjacency and backtracking in this specific kind of grid traversal problem. Still, it provides a clearer structure for paths explored and can be integrated with queue data structures for parallel processing.

The key difficulty with BFS in this problem is efficiently pruning unfruitful search branches early, which DFS inherently handles through natural backtracking.

This Python example uses a deque to handle BFS traversal. For each starting point matching the beginning of word, it attempts to queue possible paths that match subsequent characters. Although this implementation correctly attempts breadth-level searches, it lacks efficiency and effectiveness compared to DFS in terms of space and early pruning.

We can enumerate each position (i, j) in the grid as the starting point of the search, and then start a depth-first search from the starting point. If we can search to the end of the word, it means the word exists, otherwise, it means the word does not exist.

Therefore, we design a function dfs(i, j, k), which represents whether we can successfully search from the (i, j) position of the grid, starting from the kth character of the word. The execution steps of the function dfs(i, j, k) are as follows:

• If k = |word|-1, it means that we have searched to the last character of the word. At this time, we only need to judge whether the character at the (i, j) position of the grid is equal to word[k]. If they are equal, it means the word exists, otherwise, it means the word does not exist. Whether the word exists or not, there is no need to continue to search, so return the result directly.
• Otherwise, if the word[k] character is not equal to the character at the (i, j) position of the grid, it means that the search failed this time, so return false directly.
• Otherwise, we temporarily store the character at the (i, j) position of the grid in c, and then modify the character at this position to a special character '0', indicating that the character at this position has been used to prevent it from being reused in subsequent searches. Then we start from the up, down, left, and right directions of the (i, j) position to search for the k+1th character in the grid. If any direction is successful, it means the search is successful, otherwise, it means the search failed. At this time, we need to restore the character at the (i, j) position of the grid, that is, put c back to the (i, j) position (backtracking).

In the main function, we enumerate each position (i, j) in the grid as the starting point. If calling dfs(i, j, 0) returns true, it means the word exists, otherwise, it means the word does not exist, so return false.

The time complexity is O(m times n times 3^k), and the space complexity is O(min(m times n, k)). Here, m and n are the number of rows and columns of the grid, respectively; and k is the length of the string word.