<p>Given an <code>m x n</code> <code>board</code>&nbsp;of characters and a list of strings <code>words</code>, return <em>all words on the board</em>.</p>

<p>Each word must be constructed from letters of sequentially adjacent cells, where <strong>adjacent cells</strong> are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/07/search1.jpg" style="width: 322px; height: 322px;" />
<pre>
<strong>Input:</strong> board = [[&quot;o&quot;,&quot;a&quot;,&quot;a&quot;,&quot;n&quot;],[&quot;e&quot;,&quot;t&quot;,&quot;a&quot;,&quot;e&quot;],[&quot;i&quot;,&quot;h&quot;,&quot;k&quot;,&quot;r&quot;],[&quot;i&quot;,&quot;f&quot;,&quot;l&quot;,&quot;v&quot;]], words = [&quot;oath&quot;,&quot;pea&quot;,&quot;eat&quot;,&quot;rain&quot;]
<strong>Output:</strong> [&quot;eat&quot;,&quot;oath&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/07/search2.jpg" style="width: 162px; height: 162px;" />
<pre>
<strong>Input:</strong> board = [[&quot;a&quot;,&quot;b&quot;],[&quot;c&quot;,&quot;d&quot;]], words = [&quot;abcb&quot;]
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == board.length</code></li>
	<li><code>n == board[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 12</code></li>
	<li><code>board[i][j]</code> is a lowercase English letter.</li>
	<li><code>1 &lt;= words.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
	<li>All the strings of <code>words</code> are unique.</li>
</ul>

class TrieNode {
    public:
    unordered_map<char, TrieNode*> children;
    string word;

    TrieNode() : word("") {}
};

class Solution {
public:

    vector<string> ans;

    void addToTrie(string word, TrieNode *node) {
        for (int i = 0; i < word.length(); i++) {
            if (node->children.find(word[i]) != node->children.end()) {
                node = node->children[word[i]];
            } else {
                TrieNode* newNode = new TrieNode();
                node->children[word[i]] = newNode;
                node = newNode;
            }
        }
    }

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

        // Construct the trie.
        for(int i = 0; i < words.size(); i++){
            TrieNode* node = root;
            //addToTrie(words[i], node);
            for (int j = 0; j < words[i].length(); j++) {
                if (node->children.find(words[i][j]) != node->children.end()) {
                    node = node->children[words[i][j]];
                } else {
                    TrieNode* newNode = new TrieNode();
                    node->children[words[i][j]] = newNode;
                    node = newNode;
                }
            }
            node->word = words[i];
        }

        int n = board.size();
        int m = board[0].size();

        for (int i = 0; i < n; i++) {
            for (int j = 0 ; j < m; j++) {
                if (root->children.find(board[i][j]) != root->children.end()) {
                    //backtracking(i, j, root, board);
                    backtrack1(root, board, i ,j);
                }
            }
        }
        return ans;
    }

    void backtrack1(TrieNode* node, vector<vector<char>> &board, int r, int c) {

        vector<int> dx = {0, 1,  0, -1};
        vector<int> dy = {1, 0, -1,  0};

        TrieNode* current = node->children[board[r][c]];

        if (current->word != "") {
            ans.push_back(current->word);
            current->word = "";
        }


        char ch = board[r][c];
        board[r][c] = '#';

        for (int i = 0; i < 4; i++) {

            int newR = r + dx[i];
            int newC = c + dy[i];

            if (newR < 0 || newR >= board.size() || newC < 0 || newC >= board[0].size()) {
                continue;
            }

            if (current->children.find(board[newR][newC]) != current->children.end()) {
                // Next element matches with the next trie node
                backtrack1(current, board, newR, newC);
            }

        }

        board[r][c] = ch;

        if (current->children.empty()) {
             node->children.erase(board[r][c]);
        }

    }
};
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