# Approach 1: Brute Force

`class Solution(object):    def candyCrush(self, board):        """        :type board: List[List[int]]        :rtype: List[List[int]]        """        m, n = len(board), len(board[0])        def gravity():            for j in range(n):                stack = [board[i][j] for i in range(m - 1, -1, -1) if board[i][j] > 0]                stack += [0] *  (m - len(stack))                for i in range(m): board[i][j] = stack.pop()        def crush():            crush = False            for i in range(m):                for j in range(n):                    if j > 1 and board[i][j] > 0 and board[i][j] == abs(board[i][j - 1]) == abs(board[i][j - 2]):                        board[i][j - 2:j + 1] = [-abs(board[i][j]) for _ in range(3)]                        crush = True                    if i > 1 and board[i][j] != 0 and abs(board[i][j]) == abs(board[i - 1][j]) == abs(board[i - 2][j]):                        if board[i][j] > 0: board[i][j] *= -1                        if board[i - 1][j] > 0: board[i - 1][j] *= -1                        if board[i - 2][j] > 0: board[i - 2][j] *= -1                        crush = True            return crush          while crush(): gravity()        return board`

Find all the values to be deleted and make them negative and then remove all the negative values from the columns one by one

Time Complexity: O(M*N*C) where C is the number of crushing rounds req

Space Complexity: O(M*N)

# Approach 2: Rotate array to ease gravity

`class Solution(object):    def candyCrush(self, board):        """        :type board: List[List[int]]        :rtype: List[List[int]]        """        board = map(list,zip(*reversed(board)))        R, C = len(board), len(board[0])        while True:            remove = set()            for i in range(R):                for j in range(2, C):                    if board[i][j] != 0 and board[i][j] == board[i][j-1] == board[i][j-2]:                        remove |= {(i,j), (i,j-1), (i,j-2)}            for j in range(C):                for i in range(2, R):                    if board[i][j] != 0 and board[i][j] == board[i-1][j] == board[i-2][j]:                        remove |= {(i,j), (i-1,j), (i-2,j)}            if remove:                for i,j in remove: board[i][j] = 0                for i in range(R):                    count = 0                    board[i] = [x for x in board[i] if x]                    board[i].extend([0]*(C - len(board[i])))            else:                break        return list(reversed(map(list,zip(*board))))`

After iterating over all rows and columns each time to find a continuous block of candies to remove, we make them zero. Now instead of having a "gravity" function what made sense was to actually rotate the array counter clockwise and shift each of the items in the row left instead of down in a column.

Time Complexity: O(M*N*C) where C is the number of crushing rounds req

Space Complexity: O(M*N)