Candy Crush — LeetCode 723 Python Solution

MediumLeetCode PremiumArrayTwo PointersMatrixSimulation
Problem
#723
Reading time
6 min

The 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.

Example

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]]

Python solution

Python
class Solution:
    def candyCrush(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

MeasureComplexity
TimeO(m^2 \times n^2), where m and n are the number of rows and columns of the matrix, respectively
SpaceO(1) auxiliary

Pattern: Two Pointers

Use the order already in the input to discard half the search space at every step. LeetCode 723. Candy Crush is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Two Pointers.

The two pointers guide has the Python template for the pattern and the 201 LeetCode problems that use it.

Related problems

Frequently asked questions

How hard is LeetCode 723. Candy Crush?
LeetCode 723. Candy Crush is rated Medium on LeetCode.
What is the time complexity of LeetCode 723. Candy Crush?
The Python solution on this page runs in O(m^2 \times n^2), where m and n are the number of rows and columns of the matrix, respectively.
What is the space complexity of LeetCode 723. Candy Crush?
The Python solution on this page uses O(1) auxiliary space.
What topics does LeetCode 723. Candy Crush cover?
LeetCode 723. Candy Crush is tagged Array, Two Pointers, Matrix and Simulation on LeetCode.
Is LeetCode 723. Candy Crush a premium problem?
Yes. LeetCode 723. Candy Crush is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

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