> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Candy Crush Python Solution with Tests

> Tested Python solution for LeetCode 723 with 13 pytest cases. Generate a practice environment with lcpy.

LeetCode 723, [Medium](/catalog/medium). Topics: [Array](/catalog/topics/array), [Two Pointers](/catalog/topics/two-pointers), [Matrix](/catalog/topics/matrix), [Simulation](/catalog/topics/simulation). [View on LeetCode](https://leetcode.com/problems/candy-crush/description/).

Generate this problem as a practice environment: tested reference solution, 13 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
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](https://github.com/wislertt/leetcode-py/blob/main/leetcode/candy_crush/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/candy_crush/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
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) |

## Tags

[NeetCode All](/catalog/neetcode).


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