Today I have done leetcode's October LeetCoding Challenge and Word Search II with cpp.

October LeetCoding Challenge 7

Description

Word Search

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:

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

Example 2:

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

Example 3:

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

Constraints:

• m == board.length
• n = board[i].length
• 1 <= m, n <= 6
• 1 <= word.length <= 15
• board and word consists of only lowercase and uppercase English letters.

Follow up: Could you use search pruning to make your solution faster with a larger board?

Solution

struct TrieNode {
  bool end = false;
  TrieNode *child[128] = {};
};
void insert(TrieNode *root, string word) {
  TrieNode *cur = root;
  for(auto c : word) {
    if(!cur->child[c]) {
      cur->child[c] = new TrieNode();
    }
    cur = cur->child[c];
  }
  cur->end = true;
}
constexpr int moves[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
class Solution {
  vector<vector<bool>> vis;
  vector<string> words;
  string current;
  int rows;
  int cols;
  void solve(vector<vector<char>>& board, TrieNode *root, int row, int col) {
    root = root->child[board[row][col]];
    vis[row][col] = true;
    current.push_back(board[row][col]);
    if(root) {
      if(root->end) words.push_back(current);
      for(int i = 0; i < 4; ++i) {
        int newRow = row + moves[i][0];
        int newCol = col + moves[i][1];
        if(newRow < 0 || newCol < 0 || newRow >= rows || newCol >= cols) continue;
        if(vis[newRow][newCol]) continue;
        solve(board, root, newRow, newCol);
      }
    }
    vis[row][col] = false;
    current.pop_back();
  }
public:
  bool exist(vector<vector<char>>& board, string word) {
    TrieNode *root = new TrieNode();
    rows = board.size();
    cols = board.front().size();
    vis.resize(rows, vector<bool>(cols));
    insert(root, word);
    for(int i = 0; i < rows; ++i) {
      for(int j = 0; j < cols; ++j) {
        solve(board, root, i, j);
      }
    }
    return words.size();
  }
};

// Accepted
// 57/57 cases passed (312 ms)
// Your runtime beats 62.95 % of cpp submissions
// Your memory usage beats 5.08 % of cpp submissions (8.2 MB)

Word Search II

Description

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: []

Constraints:

• m == board.length
• n == board[i].length
• 1 <= m, n <= 12
• board[i][j] is a lowercase English letter.
• 1 <= words.length <= 3 * 10^4
• 1 <= words[i].length <= 10
• words[i] consists of lowercase English letters.
• All the strings of words are unique.

Solution

struct TrieNode {
  bool end = false;
  TrieNode *child[26] = {};
};
void insert(TrieNode *root, string &word) {
  TrieNode *cur = root;
  for(auto c : word) {
    c = c - 'a';
    if(!cur->child[c]) {
      cur->child[c] = new TrieNode();
    }
    cur = cur->child[c];
  }
  cur->end = true;
}
constexpr int moves[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
class Solution {
  vector<vector<bool>> vis;
  vector<string> words;
  string current;
  int rows;
  int cols;
  void solve(vector<vector<char>>& board, TrieNode *root, int row, int col) {
    root = root->child[board[row][col] - 'a'];
    vis[row][col] = true;
    current.push_back(board[row][col]);
    if(root) {
      if(root->end) words.push_back(current);
      for(int i = 0; i < 4; ++i) {
        int newRow = row + moves[i][0];
        int newCol = col + moves[i][1];
        if(newRow < 0 || newCol < 0 || newRow >= rows || newCol >= cols) continue;
        if(vis[newRow][newCol]) continue;
        solve(board, root, newRow, newCol);
      }
    }
    vis[row][col] = false;
    current.pop_back();
  }
public:
  vector<string> findWords(vector<vector<char>>& board, vector<string>& w) {
    TrieNode *root = new TrieNode();
    rows = board.size();
    cols = board.front().size();
    vis.resize(rows, vector<bool>(cols));
    for(auto &word : w) {
      insert(root, word);
    }
    for(int i = 0; i < rows; ++i) {
      for(int j = 0; j < cols; ++j) {
        solve(board, root, i, j);
      }
    }
    sort(words.begin(), words.end());
    words.resize(unique(words.begin(), words.end()) - words.begin());
    return words;
  }
};

// Accepted
// 51/51 cases passed (1948 ms)
// Your runtime beats 5.01 % of cpp submissions
// Your memory usage beats 5.12 % of cpp submissions (290.5 MB)