Candy Crush Python Solution with Tests

LeetCode 723, Medium. Topics: Array, Two Pointers, Matrix, Simulation. View on LeetCode. Generate this problem as a practice environment: tested reference solution, 13 parametrized pytest cases, and a playground notebook:

lcpy gen -n 723   # by problem number
lcpy gen -s candy_crush   # by problem name

Problem

This question is about implementing a basic elimination algorithm for Candy Crush. Given an m x n integer array board representing the grid of candy where board[i][j] represents the type of candy. A value of board[i][j] == 0 represents that the cell is empty. The given board represents the state of the game following the player’s move. Now, you need to restore the board to a stable state by crushing candies according to the following rules:

If three or more candies of the same type are adjacent vertically or horizontally, crush them all at the same time - these positions become empty.
After crushing all candies simultaneously, if an empty space on the board has candies on top of itself, then these candies will drop until they hit a candy or bottom at the same time. No new candies will drop outside the top boundary.
After the above steps, there may exist more candies that can be crushed. If so, you need to repeat the above steps.
If there does not exist more candies that can be crushed (i.e., the board is stable), then return the current board.

You need to perform the above rules until the board becomes stable, then return the stable board.

Examples

Input: board = [[110,5,112,113,114],[210,211,5,213,214],[310,311,3,313,314],[410,411,412,5,414],[5,1,512,3,3],[610,4,1,613,614],[710,1,2,713,714],[810,1,2,1,1],[1,1,2,2,2],[4,1,4,4,1014]]
Output: [[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[110,0,0,0,114],[210,0,0,0,214],[310,0,0,113,314],[410,0,0,213,414],[610,211,112,313,614],[710,311,412,613,714],[810,411,512,713,1014]]

Input: board = [[1,3,5,5,2],[3,4,3,3,1],[3,2,4,5,2],[2,4,4,5,5],[1,4,4,1,1]]
Output: [[1,3,0,0,0],[3,4,0,5,2],[3,2,0,3,1],[2,4,0,5,2],[1,4,3,1,1]]

Constraints

m == board.length
n == board[i].length
3 <= m, n <= 50
1 <= board[i][j] <= 2000

Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O((m * n)^2)
    # Space: O(1)
    def candy_crush(self, board: list[list[int]]) -> list[list[int]]:
        m, n = len(board), len(board[0])
        run = True
        while run:
            run = False
            for i in range(m):
                for j in range(2, n):
                    if board[i][j] and abs(board[i][j]) == abs(board[i][j - 1]) == abs(
                        board[i][j - 2]
                    ):
                        run = True
                        board[i][j] = board[i][j - 1] = board[i][j - 2] = -abs(board[i][j])
            for j in range(n):
                for i in range(2, m):
                    if board[i][j] and abs(board[i][j]) == abs(board[i - 1][j]) == abs(
                        board[i - 2][j]
                    ):
                        run = True
                        board[i][j] = board[i - 1][j] = board[i - 2][j] = -abs(board[i][j])
            if run:
                for j in range(n):
                    k = m - 1
                    for i in range(m - 1, -1, -1):
                        if board[i][j] > 0:
                            board[k][j] = board[i][j]
                            k -= 1
                    while k >= 0:
                        board[k][j] = 0
                        k -= 1
        return board

Complexity

Time	Space
O((m * n)^2)	O(1)

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Candy Crush Python Solution with Tests

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