Word Search II
Problem Statement :
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 * 104 1 <= words[i].length <= 10 words[i] consists of lowercase English letters. All the strings of words are unique.
Solution :
Solution in C :
int insPos = -1;
for (int i = 0; i < indexArrayLen; i++) {
if (wordsLenArray[indexArray[i]] == seedLen) {
//find word
indexArrayLen--;
RetArray[retArrayNum++] = pWords[indexArray[i]];
wordsLenArray[indexArray[i]] = 0;
//updatetargetIdxBuf
int k = 0;
int pos;
for (k = 0; k < targetIdxNum[seed[0] - 'a']; k++) {
if (targetIdxBuf[seed[0] - 'a'][k] == indexArray[i]) {
pos = k;
break;
}
}
targetIdxNum[seed[0] - 'a']--;
if (pos != targetIdxNum[seed[0] - 'a']) {
for (k = pos; k < targetIdxNum[seed[0] - 'a']; k++)
{
targetIdxBuf[seed[0] - 'a'][k] = targetIdxBuf[seed[0] - 'a'][k + 1];
}
}
insPos = i;
break;
}
}
if (indexArrayLen == 0) {
return;
}
else if(insPos != -1 && insPos < indexArrayLen && seedLen !=1 ){
for (int k = insPos; k < indexArrayLen; k++)
{
indexArray[k] = indexArray[k + 1];
}
}
trace[row][col] = 1;
// 상하좌우로 계속 이동한다.
//상
if (row - 1 >= 0 && trace[row - 1][col] != 1){
char filter = pBoard[row - 1][col];
//indexArray 중에서 filter에 맞는 것의 개수를 찾자.
seed[seedLen] = filter;
seed[seedLen + 1] = 0;
int* idxArray = (int*)malloc(indexArrayLen * sizeof(int));
int idxNum = 0;
for (int k = 0; k < indexArrayLen; k++) {
if (wordsLenArray[indexArray[k]] > seedLen) {
if (filter == pWords[indexArray[k]][seedLen]) {
idxArray[idxNum++] = indexArray[k];
}
}
}
if (idxNum > 0)
searchWord2(seed, seedLen + 1, idxArray, idxNum, row - 1, col);
free(idxArray);
}
// 하
if (row + 1 < H && trace[row + 1][col] != 1) {
char filter = pBoard[row + 1][col];
//indexArray 중에서 filter에 맞는 것의 개수를 찾자.
seed[seedLen] = filter;
seed[seedLen + 1] = 0;
int* idxArray = (int*)malloc(indexArrayLen * sizeof(int));
int idxNum = 0;
for (int k = 0; k < indexArrayLen; k++) {
if (wordsLenArray[indexArray[k]] > seedLen) {
if (filter == pWords[indexArray[k]][seedLen]) {
idxArray[idxNum++] = indexArray[k];
}
}
}
if(idxNum > 0)
searchWord2(seed, seedLen + 1, idxArray, idxNum, row + 1, col);
free(idxArray);
}
// 좌
if (col - 1 >= 0 && trace[row][col - 1] != 1) {
char filter = pBoard[row][col -1];
//indexArray 중에서 filter에 맞는 것의 개수를 찾자.
seed[seedLen] = filter;
seed[seedLen + 1] = 0;
int* idxArray = (int*)malloc(indexArrayLen * sizeof(int));
int idxNum = 0;
for (int k = 0; k < indexArrayLen; k++) {
if (wordsLenArray[indexArray[k]] > seedLen) {
if (filter == pWords[indexArray[k]][seedLen]) {
idxArray[idxNum++] = indexArray[k];
}
}
}
if (idxNum > 0)
searchWord2(seed, seedLen + 1, idxArray, idxNum, row, col -1);
free(idxArray);
}
//우
if (col + 1 < W && trace[row][col + 1] != 1) {
char filter = pBoard[row][col+1];
//indexArray 중에서 filter에 맞는 것의 개수를 찾자.
seed[seedLen] = filter;
seed[seedLen + 1] = 0;
int* idxArray = (int*)malloc(indexArrayLen * sizeof(int));
int idxNum = 0;
for (int k = 0; k < indexArrayLen; k++) {
if (wordsLenArray[indexArray[k]] > seedLen) {
if (filter == pWords[indexArray[k]][seedLen]) {
idxArray[idxNum++] = indexArray[k];
}
}
}
if (idxNum > 0)
searchWord2(seed, seedLen + 1, idxArray, idxNum, row, col +1);
free(idxArray);
}
trace[row][col] = 0;
Solution in C++ :
class Solution
{
public:
bool dfs(int ind, int indi, int indj, vector<vector < char>> &board, string &search, int &row, int &col)
{
if (ind == search.size())
return true;
if (indi >= 0 and indi < row and indj >= 0 and indj < col)
{
if (board[indi][indj] != search[ind])
return false;
char originalchar = board[indi][indj];
board[indi][indj] = '$';
bool ans = dfs(ind + 1, indi + 1, indj, board, search, row, col) ||
dfs(ind + 1, indi - 1, indj, board, search, row, col) ||
dfs(ind + 1, indi, indj + 1, board, search, row, col) ||
dfs(ind + 1, indi, indj - 1, board, search, row, col);
board[indi][indj] = originalchar;
return ans;
}
else
return false;
}
vector<string> findWords(vector<vector < char>> &board, vector< string > &words)
{
int row = board.size();
int col = board[0].size();
vector<string> ans;
vector<vector<pair<int, int>>> vp(27);
for (int i = 0; i < row; i++)
{
for (int j = 0; j < col; j++)
vp[board[i][j] - 'a'].push_back({ i, j });
}
int starta = 0;
int enda = 0;
for (auto ele: words)
{
if (ele[0] == 'a')
starta++;
if (ele[ele.size() - 1] == 'a')
enda++;
}
bool reversehaikya = false;
if (starta > enda)
{
reversehaikya = true;
for (int i = 0; i < words.size(); i++)
reverse(words[i].begin(), words[i].end());
}
for (auto search: words)
{
bool flag = false;
for (auto ele: vp[search[0] - 'a'])
{
flag = dfs(0, ele.first, ele.second, board, search, row, col);
if (flag)
{
if (reversehaikya)
reverse(search.begin(), search.end());
ans.push_back(search);
break;
}
}
}
return ans;
}
};
Solution in Java :
class Solution {
public boolean dfs(int ind , int indi , int indj , char[][] board , String search , int row , int col){
if(ind == search.length()){
return true;
}
if(indi >= 0 && indi < row && indj>=0 && indj<col){
if(board[indi][indj] != search.charAt(ind)){
return false;
}
char originalchar = board[indi][indj];
board[indi][indj] = '$';
boolean ans = dfs(ind+1 , indi+1 , indj , board , search , row , col) ||
dfs(ind+1 , indi-1 , indj , board , search , row , col) ||
dfs(ind+1 , indi , indj+1 , board , search , row , col) ||
dfs(ind+1 , indi , indj-1 , board , search , row , col);
board[indi][indj] = originalchar;
return ans;
}
else
return false;
}
public List<String> findWords(char[][] board, String[] words) {
int row = board.length;
int col = board[0].length;
List<String> ans = new ArrayList<>();
List<List<Pair<Integer , Integer>>> v = new ArrayList<>();
for(int i=0 ; i<26 ; i++){
v.add(new ArrayList<>());
}
for(int i=0 ; i<row ; i++){
for(int j=0; j<col ; j++){
v.get(board[i][j] - 'a').add(new Pair<>(i , j));
}
}
int start = 0 , end = 0;
for(String word : words){
if(word.charAt(0) == 'a')
start++;
if(word.charAt(word.length()-1) == 'a')
end++;
}
boolean reversehaikya = false;
if(start > end){
for(int i=0 ; i<words.length; i++){
reversehaikya = true;
words[i] = new StringBuilder(words[i]).reverse().toString();
}
}
for(String search : words){
boolean flag = false;
for(Pair<Integer , Integer> ele : v.get(search.charAt(0) - 'a')){
flag = dfs(0 , ele.getKey() , ele.getValue() , board , search , row , col);
if(flag){
if(reversehaikya)
search = new StringBuilder(search).reverse().toString();
ans.add(search);
break;
}
}
}
return ans;
}
}
Solution in Python :
class Solution:
def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
m = len(board)
n = len(board[0])
res = []
d = [[0, 1], [0, -1], [1, 0], [-1, 0]]
ref = set()
for i in range(m):
for j in range(n-1):
ref.add(board[i][j] + board[i][j+1])
for j in range(n):
for i in range(m-1):
ref.add(board[i][j] + board[i+1][j])
for word in words:
f = True
for i in range(len(word)-1):
if word[i:i+2] not in ref and word[i+1] + word[i] not in ref:
f = False
break
if not f:
continue
if self.findWord(word, m, n, board, d):
res.append(word)
return res
def findWord(self, word, m, n, board, d) -> bool:
if word[:4] == word[0] * 4:
word = ''.join([c for c in reversed(word)])
starts = []
stack = []
visited = set()
for i in range(m):
for j in range(n):
if board[i][j] == word[0]:
if len(word) == 1:
return True
starts.append((i, j))
for start in starts:
stack.append(start)
visited.add((start, ))
l = 1
while stack != [] and l < len(word):
x, y = stack[-1]
for dxy in d:
nx, ny = x + dxy[0], y + dxy[1]
if 0 <= nx < m and 0 <= ny < n:
if board[nx][ny] == word[l]:
if (nx, ny) not in stack and tuple(stack) + ((nx, ny),) not in visited:
stack.append((nx, ny))
visited.add(tuple(stack))
l += 1
if l == len(word):
return True
break
else:
stack.pop()
l -= 1
else:
return False
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