Problem: Time taken by tomatoes to rot in JavaScript

Problem

We are required to write a JavaScript function that takes in a 2-D array of numbers, arr, as the only argument.

The numbers in the array can be −

• the value 0 which represents an empty cell;

• the value 1 which represents a fresh tomato;

• the value 2 which represents a rotten tomato.

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

Our function is supposed to return the minimum number of minutes that must elapse until no cell has a fresh tomato. If this is impossible, we should return -1 instead.

For example, if the input to the function is −

const arr = [
   [2, 1, 1],
   [1, 1, 0],
   [0, 1, 1]
];

Then the output should be −

const output = 4;

Output Explanation

[
[2, 2, 1],
[2, 1, 0],
[0, 1, 1]
]

[
[2, 2, 2],
[2, 2, 0],
[0, 1, 1]
]

[
[2, 2, 2],
[2, 2, 0],
[0, 2, 1]
]

[
[2, 2, 2],
[2, 2, 0],
[0, 2, 2]
]

Example

The code for this will be −

 Live Demo

const arr = [
   [2, 1, 1],
   [1, 1, 0],
   [0, 1, 1]
];
const timeToRot = (arr = []) => {
   let fresh = 0;
   let count = -1;
   let curr = [];
   for(let i = 0; i < arr.length; i++){
      for(let j = 0; j < arr[i].length; j++){
         if(arr[i][j] === 1){
            fresh += 1;
         };
         if(arr[i][j] === 2){
            curr.push([i, j]);
         };
      };
   };
   if(!fresh){
      return 0;
   };
   while(curr.length > 0){
      count += 1;
      const next = [];
      const rotten = (i, j) => {
         arr[i][j] = 2
         next.push([i, j])
         fresh -= 1
      };
      for(const [i, j] of curr){
         if (arr[i - 1] && arr[i - 1][j] === 1) {
            rotten(i - 1, j);
         };
         if (arr[i + 1] && arr[i + 1][j] === 1) {
            rotten(i + 1, j);
         };
         if (arr[i][j - 1] === 1) {
            rotten(i, j - 1);
         };
         if (arr[i][j + 1] === 1) {
            rotten(i, j + 1);
         };
      }
      curr = next
   };
   return fresh === 0 ? count : -1;
};
console.log(timeToRot(arr));

Output

And the output in the console will be −

4