Rotting Oranges

mediumcodingAlgorithmsBFSMatrix~30 min

Asked at Google, Meta

Overview

In an m × n grid, each cell can be: 0 (empty), 1 (fresh orange), or 2 (rotten orange). Every minute, fresh oranges adjacent (4-directionally) to rotten ones become rotten. Return the minimum minutes until no fresh orange remains, or -1 if impossible.

Constraints

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

Examples

orangesRotting([[2,1,1],[1,1,0],[0,1,1]]);
// => 4

orangesRotting([[2,1,1],[0,1,1],[1,0,1]]);
// => -1  (bottom-left fresh orange is unreachable)

Solution

Reveal solution

function orangesRotting(grid) {
  const m = grid.length, n = grid[0].length;
  const queue = [];
  let fresh = 0;
  for (let i = 0; i < m; i++)
    for (let j = 0; j < n; j++) {
      if (grid[i][j] === 2) queue.push([i, j]);
      else if (grid[i][j] === 1) fresh++;
    }
  if (fresh === 0) return 0;
  const dirs = [[1,0],[-1,0],[0,1],[0,-1]];
  let minutes = 0, idx = 0;
  while (idx < queue.length) {
    const size = queue.length - idx;
    let rotted = false;
    for (let s = 0; s < size; s++) {
      const [r, c] = queue[idx++];
      for (const [dr, dc] of dirs) {
        const nr = r+dr, nc = c+dc;
        if (nr >= 0 && nr < m && nc >= 0 && nc < n && grid[nr][nc] === 1) {
          grid[nr][nc] = 2;
          fresh--;
          queue.push([nr, nc]);
          rotted = true;
        }
      }
    }
    if (rotted) minutes++;
  }
  return fresh === 0 ? minutes : -1;
}

Resources

Rotting Oranges — LeetCode

rotting-oranges.js

Rotting Oranges

mediumcodingAlgorithmsBFS

Overview

In an m × n grid, each cell can be: 0 (empty), 1 (fresh orange), or 2 (rotten orange). Every minute, fresh oranges adjacent (4-directionally) to rotten ones become rotten. Return the minimum minutes until no fresh orange remains, or -1 if impossible.

Constraints

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

Examples

orangesRotting([[2,1,1],[1,1,0],[0,1,1]]);
// => 4

orangesRotting([[2,1,1],[0,1,1],[1,0,1]]);
// => -1  (bottom-left fresh orange is unreachable)

Solution

Reveal solution

function orangesRotting(grid) {
  const m = grid.length, n = grid[0].length;
  const queue = [];
  let fresh = 0;
  for (let i = 0; i < m; i++)
    for (let j = 0; j < n; j++) {
      if (grid[i][j] === 2) queue.push([i, j]);
      else if (grid[i][j] === 1) fresh++;
    }
  if (fresh === 0) return 0;
  const dirs = [[1,0],[-1,0],[0,1],[0,-1]];
  let minutes = 0, idx = 0;
  while (idx < queue.length) {
    const size = queue.length - idx;
    let rotted = false;
    for (let s = 0; s < size; s++) {
      const [r, c] = queue[idx++];
      for (const [dr, dc] of dirs) {
        const nr = r+dr, nc = c+dc;
        if (nr >= 0 && nr < m && nc >= 0 && nc < n && grid[nr][nc] === 1) {
          grid[nr][nc] = 2;
          fresh--;
          queue.push([nr, nc]);
          rotted = true;
        }
      }
    }
    if (rotted) minutes++;
  }
  return fresh === 0 ? minutes : -1;
}

Resources

Rotting Oranges — LeetCode

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