31. Word Search II

Given an m x n board of characters and a list of strings words, return all words on the board.

Each word must 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 in a word.

Example 1:

Input: board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]
Output: ["eat","oath"]

Example 2:

Input: board = [["a","b"],["c","d"]], words = ["abcb"]
Output: []

Solution: (Backtracking)

class Solution
{
public:
    bool dfs(vector<vector<char>> &board, int i, int j, string word, int pos)
    {
        int n = board.size();
        int m = board[0].size();

        if (pos == word.length())
        {
            return true;
        }

        if (i >= 0 && j >= 0 && i < n && j < m && word[pos] == board[i][j])
        {
            
            char temp = board[i][j];
            board[i][j] = ' ';

            if (dfs(board, i + 1, j, word, pos + 1))
            {
                 board[i][j] = temp;
                return true;
            }

            if (dfs(board, i, j + 1, word, pos + 1))
            {
                 board[i][j] = temp;
                return true;
            }

            if (dfs(board, i - 1, j, word, pos + 1))
            {
                 board[i][j] = temp;
                return true;
            }

            if (dfs(board, i, j - 1, word, pos + 1))
            {
                 board[i][j] = temp;
                return true;
            }
            
            board[i][j] = temp;
        }

        return false;
    }

    vector<string> findWords(vector<vector<char>> &board, vector<string> &words)
    {
        vector<string> res;

        for (int k = 0; k < words.size(); k++)
        {
            bool flag = false;
            for (int i = 0; i < board.size(); i++)
            {
                for (int j = 0; j < board[i].size(); j++)
                {
                    if (words[k][0] == board[i][j])
                    {
                        if (dfs(board, i, j, words[k], 0))
                        {
                            flag = true;
                            res.push_back(words[k]);
                            break;
                        }
                    }
                }

                if (flag)
                {
                    break;
                }
            }
        }

        return res;
    }
};

Time Complexity: (nm * nm * numWords)

Solution: (Trie)

class Solution
{
public:
    struct TrieNode
    {
        char val;
        int endHere;
        string word;
        TrieNode *child[26];
    };

    TrieNode *createNode(char ch)
    {
        TrieNode *temp = new TrieNode;
        temp->val = ch;
        temp->endHere = 0;
        temp->word = "";
        for (int i = 0; i < 26; i++)
        {
            temp->child[i] = NULL;
        }

        return temp;
    }

    vector<string> res;

    void dfs(vector<vector<char>> &board, TrieNode *root, int i, int j)
    {
        int n = board.size();
        int m = board[0].size();

        if (i < 0 || j < 0 || i >= n || j >= m)
        {
            return;
        }

        if (board[i][j] == '*')
        {
            return;
        }

        int idx = board[i][j] - 97;

        root = root->child[idx];

        if (root == NULL || root->val != board[i][j])
        {
            return;
        }

        if (root->endHere > 0)
        {
            res.push_back(root->word);
            root->endHere--;
        }

        char temp = board[i][j];
        board[i][j] = '*';

        dfs(board, root, i + 1, j);
        dfs(board, root, i - 1, j);
        dfs(board, root, i, j + 1);
        dfs(board, root, i, j - 1);

        board[i][j] = temp;
        return;
    }

    void insert(string word, TrieNode *root)
    {
        TrieNode *p = root;
        for (int i = 0; i < word.length(); i++)
        {
            int node = word[i] - 97;
            if (!p->child[node])
            {
                p->child[node] = createNode(word[i]);
            }
            p = p->child[node];
        }
        p->endHere++;
        p->word = word;
    }

    vector<string> findWords(vector<vector<char>> &board, vector<string> &words)
    {
        TrieNode *root = createNode('/');

        for (int i = 0; i < words.size(); i++)
        {
            insert(words[i], root);
        }

        for (int i = 0; i < board.size(); i++)
        {
            for (int j = 0; j < board[0].size(); j++)
            {
                int ch = board[i][j] - 97;
                if (root->child[ch])
                {
                    dfs(board, root, i, j);
                }
            }
        }
        return res;
    }
};

Time Complexity: (nm * nm)