Question

Given a 2D board and a list of words from the dictionary, find all words in the board.

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

Difficulty:Hard

Category:Backtracking, Trie

Analyze

This question is similar with the Leetcode 79: Word Search, I reused the wordsearch recursion function in that problem and change the loop in the findWords to search for each words. As a result, the runtime around 800ms.

Solution

// Runtime: 780ms
class Solution {
 public:
  vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
    unordered_set<string> rec;
    vector<string> ans;
    if (board.empty()) return {};
    int m = board.size(), n = board[0].size();
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        for (string& word : words) {
          if (rec.count(word)) continue;
          if (wordsearch(board, word, 0, i, j) ) {
            rec.insert(word);
            ans.emplace_back(word);
          }
        }
      }
    }
    return ans;
  }

 private:
  bool wordsearch(vector<vector<char>>& b, string& w, int d, int x, int y) { 
    if (x < 0 || y < 0 || x == b.size() || y == b[0].size() || w[d] != b[x][y])
      return false;
    if (d == w.length() - 1) return true;
    char cur = b[x][y];
    b[x][y] = 0;
    bool res = wordsearch(b, w, d + 1, x + 1, y) 
              || wordsearch(b, w, d + 1, x, y + 1) 
              || wordsearch(b, w, d + 1, x - 1, y) 
              || wordsearch(b, w, d + 1, x, y - 1);
    b[x][y] = cur;
    return res;
  }
};