Rotting Oranges

Problem Description

You are given an m x n grid where each cell can have one of three values:

• 0 representing an empty cell,
• 1 representing a fresh orange, or
• 2 representing a rotten orange.

Every minute, any fresh orange that is 4-directionally adjacent to a rotten orange becomes rotten.

Return the minimum number of minutes that must elapse until no cell has a fresh orange. If this is impossible, return -1.

Example 1:

Input: grid = [[2,1,1],[1,1,0],[0,1,1]]
Output: 4

Example 2:

Input: grid = [[2,1,1],[0,1,1],[1,0,1]]
Output: -1
Explanation: The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.

Example 3:

Input: grid = [[0,2]]
Output: 0
Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 10
• grid[i][j] is 0, 1, or 2.

Solution (JavaScript)

/**
 * @param {number[][]} grid
 * @return {number}
 */
var orangesRotting = function(grid) {
    const flatten = (a) => a.reduce((p, c) => p.concat(c));
    const by2 = (flat, idx, wth) => {
        return flat[idx-1] === 2 || flat[idx+1] === 2 || flat[idx-wth] === 2 || flat[idx+wth] === 2;
    }
    
    const h = grid.length, w = grid[0].length;
    let loop = h*w;
    let days = 0, tmp = 0;
    
    while(loop){
        const flat = flatten(grid.map((a) => a.concat([0])));
        for(let i = 0; i < h; i++){
            for(let j = 0; j < w; j++){
                if(grid[i][j] === 1 && by2(flat, (w+1)*i+j, w+1)){
                    tmp = tmp | 1;
                    grid[i][j] = 2;
                    console.log('tada', i, j);
                }
            }
        }
        if(tmp){
            days += tmp;
            console.log('tmp', tmp);
        } else {
            console.log('break');
            break;
        }
        tmp = 0;
        loop -= 1;
    }
    
    if(flatten(grid).includes(1)){
        return -1;
    } else {
        return days;
    }
};