Problem Statement

Given a 2D integer array board representing the grid of candy, different positive integers, each in board[i][j] represent different types of candies. A value of board[i][j]= 0 represents that the cell at position (i, j) 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:

  1. 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.
  2. 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.)
  3. After the above steps, there may exist more candies that can be crushed. If so, you need to repeat the above steps.
  4. If there does not exist more candies that can be crushed (ie. the board is stable), then return the current board. You need to perform the above rules until the board becomes stable, then return the current board.

class Solution {
    public int[][] candyCrush(int[][] board) {
    	// Your Code Here
    }
}

Test Cases

Test Case 1

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]]

Follow Up Questions To Think About

  • Suppose the array was instead a linked list. How would this implementation change? What would be the consequences?
  • What is the runtime of your approach?

Click Below to see the Solution

Algorithm

The idea is to traverse the entire matrix with a while loop to remove crush until no crush can be found. See the below code comments for a more concrete algorithm.

Code


class Solution {
    public int[][] candyCrush(int[][] board) {
        int N = board.length;
        int M = board[0].length;
        // found will determine if the board has any potential changes to make
        boolean found = true;
        while (found) {
            found = false;
            for (int i = 0; i < N; i++) {
                for (int j = 0; j < M; j++) {
                    int val = Math.abs(board[i][j]);
                    if (val == 0) continue;
                    // horizontally (to the right) checking if has three at the same time
                    if (((j <= M - 3) && (Math.abs(board[i][j + 1])) == val) 
                        && (Math.abs(board[i][j + 2]) == val)) {
                        found = true;
                        int ind = j;
                        // marking them negative values
                        while ((ind < M) && (Math.abs(board[i][ind]) == val)) {
                            board[i][ind++] = -val;
                        }
                    }
                    // vertically (downwards) checking if has three at the same time
                    if (((i <= N - 3) && (Math.abs(board[i + 1][j]) == val)) 
                        && (Math.abs(board[i + 2][j]) == val)) {
                        found = true;
                        int ind = i;
                        // marking them negative values
                        while ((ind < N) && (Math.abs(board[ind][j]) == val)) {
                            board[ind++][j] = -val; 
                        }
                    }
                }
            }
            // if we have eliminate any blocks by marking them negative, we will finally adjust the board
            // therefore one iteration
            if (found) { // move positive values to the bottom, then set the rest to 0
                for (int j = 0; j < M; j++) {
                    int storeInd = N - 1;
                    for (int i = N - 1; i >= 0; i--) {
                        if (board[i][j] > 0) {
                            board[storeInd--][j] = board[i][j];
                        }
                    }
                    for (int k = storeInd; k >= 0; k--){
                        board[k][j] = 0;
                    }
                }
            }
        }
        return board;
    }
}

Runtime Analysis

The computational complexity is as the program keeps looping through the matrix until it can no longer find a candy crush.

For each traversal of the matrix, we only check two directions, rightward and downward. There is no need to check upward and leftward because they would have been checked from previous cells.

For each cell, we check if there are at least three candies of the same type to the right of the current or downward cells, set them all to their negative values marking them as crushes.

After each traversal, we need to remove all those negative values by setting them to 0, then let the rest drop down to change to their correct position.