You are given an m x ngrid 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.
Pseudocode
- bfs to look for fresh oranges starting from rotten oranges, moving in 4 directions
- reassign values of fresh oranges, to distance from rotten oranges
Solution
var orangesRotting = function (grid) {
if (!grid || grid.length === 0) {
return 0;
}
const row = grid.length;
const column = grid[0].length;
const queue = [];
let countFresh = 0;
let minutes = 0;
let dirs = [
[1, 0],
[-1, 0],
[0, 1],
[0, -1],
];
// put the position of all rotten oranges in queue
// count the number of fresh oranges
for (let i = 0; i < row; i++) {
for (let j = 0; j < column; j++) {
if (grid[i][j] === 2) {
queue.push([i, j]);
}
if (grid[i][j] === 1) {
countFresh++;
}
}
}
// if count of fresh oranges is zero, return zero
if (countFresh === 0) {
return 0;
}
// bfs starting from initially rotten oranges
while (queue.length !== 0 && countFresh) {
minutes++;
let size = queue.length; // set the size before hand,
for (let i = 0; i < size; i++) {
// if i is limited by queue.length i.e. i < queue.length, queue.length will be vary dynamically
const point = queue.shift();
for (const dir of dirs) {
const x = point[0] + dir[0];
const y = point[1] + dir[1];
// set when to continue, out of bounds, already rotten or empty cell
if (
x < 0 ||
y < 0 ||
x >= row ||
y >= column ||
grid[x][y] === 2 ||
grid[x][y] === 0
) {
continue;
}
// if code reaches here, orange is fresh, mark is as rotten
grid[x][y] = 2;
// book keeping, if not all fresh oranges are rotten, return -1
countFresh--;
// push this the coords of this newly rotten orange into queue to look for next fresh orange
queue.push([x, y]);
}
}
}
// count - 1 because minute is 0 based index
return countFresh === 0 ? minutes : -1;
};
