079 Word Search.py

'''
Given a 2D board and a word, find if the word exists in the grid.

The word can 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.

For example,
Given board =

[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.
'''

from collections import defaultdict


class Solution(object):
    def exist(self, board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        if self._hasEnoughCharacters(board, word):
            m = len(board)
            n = len(board[0])
            for i in range(m):
                for j in range(n):
                    if self._exist(board, i, j, m, n, word):
                        return True
            return False
        else:
            return False

    def _exist(self, board, i, j, m, n, word):
        if len(word) == 0:
            return True
        if i < 0 or i >= m or j < 0 or j >= n or board[i][j] != word[0]:
            return False
        temp = board[i][j]
        board[i][j] = "."
        next_target = word[1:]
        next_result = self._exist(board, i - 1, j, m, n, next_target) \
                      or self._exist(board, i + 1, j, m, n, next_target) \
                      or self._exist(board, i, j - 1, m, n, next_target) \
                      or self._exist(board, i, j + 1, m, n, next_target)
        board[i][j] = temp
        return next_result

    def _hasEnoughCharacters(self, board, word):
        character_counts = defaultdict(int)
        for ch in word:
            character_counts[ch] += 1
        return all(sum(map(lambda line: line.count(ch), board)) >= count for ch, count in character_counts.items())


if __name__ == "__main__":
    assert Solution().exist([
        ['A', 'B', 'C', 'E'],
        ['S', 'F', 'C', 'S'],
        ['A', 'D', 'E', 'E']
    ], "ABCCED") == True
    assert Solution().exist([
        ['A', 'B', 'C', 'E'],
        ['S', 'F', 'C', 'S'],
        ['A', 'D', 'E', 'E']
    ], "SEE") == True
    assert Solution().exist([
        ['A', 'B', 'C', 'E'],
        ['S', 'F', 'C', 'S'],
        ['A', 'D', 'E', 'E']
    ], "ABCB") == False