<p>You are given an <code>m x n</code> matrix <code>board</code>, representing the<strong> current </strong>state of a crossword puzzle. The crossword contains lowercase English letters (from solved words), <code>&#39; &#39;</code> to represent any <strong>empty </strong>cells, and <code>&#39;#&#39;</code> to represent any <strong>blocked</strong> cells.</p>

<p>A word can be placed<strong> horizontally</strong> (left to right <strong>or</strong> right to left) or <strong>vertically</strong> (top to bottom <strong>or</strong> bottom to top) in the board if:</p>

<ul>
	<li>It does not occupy a cell containing the character <code>&#39;#&#39;</code>.</li>
	<li>The cell each letter is placed in must either be <code>&#39; &#39;</code> (empty) or <strong>match</strong> the letter already on the <code>board</code>.</li>
	<li>There must not be any empty cells <code>&#39; &#39;</code> or other lowercase letters <strong>directly left or right</strong><strong> </strong>of the word if the word was placed <strong>horizontally</strong>.</li>
	<li>There must not be any empty cells <code>&#39; &#39;</code> or other lowercase letters <strong>directly above or below</strong> the word if the word was placed <strong>vertically</strong>.</li>
</ul>

<p>Given a string <code>word</code>, return <code>true</code><em> if </em><code>word</code><em> can be placed in </em><code>board</code><em>, or </em><code>false</code><em> <strong>otherwise</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/10/04/crossword-ex1-1.png" style="width: 478px; height: 180px;" />
<pre>
<strong>Input:</strong> board = [[&quot;#&quot;, &quot; &quot;, &quot;#&quot;], [&quot; &quot;, &quot; &quot;, &quot;#&quot;], [&quot;#&quot;, &quot;c&quot;, &quot; &quot;]], word = &quot;abc&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> The word &quot;abc&quot; can be placed as shown above (top to bottom).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/10/04/crossword-ex2-1.png" style="width: 180px; height: 180px;" />
<pre>
<strong>Input:</strong> board = [[&quot; &quot;, &quot;#&quot;, &quot;a&quot;], [&quot; &quot;, &quot;#&quot;, &quot;c&quot;], [&quot; &quot;, &quot;#&quot;, &quot;a&quot;]], word = &quot;ac&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong> It is impossible to place the word because there will always be a space/letter above or below it.</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/10/04/crossword-ex3-1.png" style="width: 478px; height: 180px;" />
<pre>
<strong>Input:</strong> board = [[&quot;#&quot;, &quot; &quot;, &quot;#&quot;], [&quot; &quot;, &quot; &quot;, &quot;#&quot;], [&quot;#&quot;, &quot; &quot;, &quot;c&quot;]], word = &quot;ca&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> The word &quot;ca&quot; can be placed as shown above (right to left). 
</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;= 2 * 10<sup>5</sup></code></li>
	<li><code>board[i][j]</code> will be <code>&#39; &#39;</code>, <code>&#39;#&#39;</code>, or a lowercase English letter.</li>
	<li><code>1 &lt;= word.length &lt;= max(m, n)</code></li>
	<li><code>word</code> will contain only lowercase English letters.</li>
</ul>

class Solution:
    def placeWordInCrossword(self, board: List[List[str]], word: str) -> bool:
        """
        Input: list of list of str, str
        Output: bool
        
        Constraint:
        1. m, n > 0, range?
        2. word > 0?
        3. english letter, " ", "#"?
        
        Alg:
        Brute Force:
        
        m rows
        n cols
        
        for each row or col
        check string matching
        
        Row:
        0 [X O X], find the white and char
        1 [O O X]
        2 [X c O]
        
        Col:
        0 [X O X]
        1 [O O c]
        2 [X X O]
        
        TC: O(M + N)
        SC: O(M * N)
        """
        
        rows_candidate = []
        cols_candidate = []
        
        m = len(board)
        n = len(board[0])
        len_word = len(word)
        
        # 1. Find the row cand.
        for row in range(m):
            tmp = {}
            start = 0
            
            for idx in range(n):
                if (board[row][idx] != "#"):
                    if (board[row][idx] != " "):
                        tmp[start] = board[row][idx]
                        
                    start += 1
                elif (start != 0):
                    if (start == len_word):
                        rows_candidate.append(tmp)
                    start = 0
                    tmp = {}
                    
            if (start == len_word):
                rows_candidate.append(tmp)
                
            tmp = {}
            start = 0
                
            for idx in range(n - 1, -1, -1):
                if (board[row][idx] != "#"):
                    if (board[row][idx] != " "):
                        tmp[start] = board[row][idx]
                        
                    start += 1
                elif (start != 0):
                    if (start == len_word):
                        rows_candidate.append(tmp)
                    start = 0
                    tmp = {}
                    
            if (start == len_word):
                rows_candidate.append(tmp)
                    
        # 2. Find the col cand.
        for col in range(n):
            tmp = {}
            start = 0
            
            for idx in range(m):
                if (board[idx][col] != "#"):
                    if (board[idx][col] != " "):
                        tmp[start] = board[idx][col]
                        
                    start += 1
                elif (start != 0):
                    if (start == len_word):
                        cols_candidate.append(tmp)
                    start = 0
                    tmp = {}
                    
            if (start == len_word):
                cols_candidate.append(tmp)
                
            tmp = {}
            start = 0
            
            for idx in range(m - 1, -1, -1):
                if (board[idx][col] != "#"):
                    if (board[idx][col] != " "):
                        tmp[start] = board[idx][col]
                        
                    start += 1
                elif (start != 0):
                    if (start == len_word):
                        cols_candidate.append(tmp)
                    start = 0
                    tmp = {}
                    
            if (start == len_word):
                cols_candidate.append(tmp)
                
        # 3. Check row cand
        match_map = {}
        
        for idx, c in enumerate(word):
            match_map[idx] = c
        
        for row_candidate in rows_candidate:
            num_match = 0
            for k, v in row_candidate.items():
                if (match_map[k] == v):
                    num_match += 1
            
            if (num_match == len(row_candidate)):
                return True
        
        # Check col cand
        for col_candidate in cols_candidate:
            num_match = 0
            for k, v in col_candidate.items():
                if (match_map[k] == v):
                    num_match += 1
            
            if (num_match == len(col_candidate)):
                return True
            
        
        return False
        
        
        
        
        
        
        
        
        
        
        
        
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