Add solution #994

Author: JoshCrozier

README.md

@@ -164,6 +164,7 @@
 977|[Squares of a Sorted Array](./0977-squares-of-a-sorted-array.js)|Easy|
 985|[Sum of Even Numbers After Queries](./0985-sum-of-even-numbers-after-queries.js)|Easy|
 989|[Add to Array-Form of Integer](./0989-add-to-array-form-of-integer.js)|Easy|
+994|[Rotting Oranges](./0994-rotting-oranges.js)|Medium|
 997|[Find the Town Judge](./0997-find-the-town-judge.js)|Easy|
 1009|[Complement of Base 10 Integer](./1009-complement-of-base-10-integer.js)|Easy|
 1010|[Pairs of Songs With Total Durations Divisible by 60](./1010-pairs-of-songs-with-total-durations-divisible-by-60.js)|Medium|

solutions/0994-rotting-oranges.js

@@ -0,0 +1,50 @@
+/**
+ * 994. Rotting Oranges
+ * https://leetcode.com/problems/rotting-oranges/
+ * Difficulty: Medium
+ *
+ * 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.
+ */
+
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var orangesRotting = function(grid) {
+  const queue = [];
+  let count = 0;
+
+  for (let i = 0; i < grid.length; i++) {
+    for (let j = 0; j < grid[0].length; j++) {
+      if (grid[i][j] === 2) {
+        queue.push([i, j, 0]);
+      }
+    }
+  }
+
+  while (queue.length) {
+    const [i, j, k] = queue.shift();
+    [[-1, 0], [1, 0], [0, -1], [0, 1]].forEach(p => {
+      const [x, y, z] = [i + p[0], j + p[1], k + 1];
+      if (x > -1 && x < grid.length && y > -1 && y < grid[0].length) {
+        if (grid[x][y] === 1) {
+          grid[x][y] = 2;
+          count = Math.max(count, z);
+          queue.push([x, y, z]);
+        }
+      }
+    });
+  }
+
+  const isIncomplete = grid.some(row => row.some(n => n === 1));
+  return isIncomplete ? -1 : count;
+};