994. Rotting Oranges.md

994. Rotting Oranges

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leetcode Daily Challenge on August 9th, 2020.

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Difficulty : Medium

Related Topics : BFS

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In a given grid, each cell can have one of three values:

• the value 0 representing an empty cell;
• the value 1 representing a fresh orange;
• the value 2 representing a rotten orange.

Every minute, any fresh orange that is adjacent (4-directionally) 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 instead.

Example 1:

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

Example 2:

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

Note:

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

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Solution

• mine
  • Java
    • BFS Runtime: 2 ms, faster than 97.60%, Memory Usage: 39 MB, less than 85.21% of Java online submissions

      // O(r * c)time
      // O(Max(len(list)))space
      public int orangesRotting(int[][] grid) {
          LinkedList<int[]> list = new LinkedList<>();
          int r = grid.length, c = grid[0].length;
          int freshCount = 0;
          for (int i = 0; i < r; i++) {
              for (int j = 0; j < c; j++) {
                  if (grid[i][j] == 2) list.add(new int[]{i, j});
                  if (grid[i][j] == 1) freshCount++;
              }
          }
          int res = 0;
          while (!list.isEmpty() && freshCount != 0) {
              res++;
              int size = list.size();
              while (size > 0) {
                  size--;
                  int[] t = list.removeFirst();
                  if (t[0] > 0 && grid[t[0] - 1][t[1]] == 1) {
                      freshCount--;
                      grid[t[0] - 1][t[1]] = 2;
                      list.add(new int[]{t[0] - 1, t[1]});
                  }
                  if (t[0] + 1 < r && grid[t[0] + 1][t[1]] == 1) {
                      freshCount--;
                      grid[t[0] + 1][t[1]] = 2;
                      list.add(new int[]{t[0] + 1, t[1]});
                  }
                  if (t[1] > 0 && grid[t[0]][t[1] - 1] == 1) {
                      freshCount--;
                      grid[t[0]][t[1] - 1] = 2;
                      list.add(new int[]{t[0], t[1] - 1});
                  }
                  if (t[1] + 1 < c && grid[t[0]][t[1] + 1] == 1) {
                      freshCount--;
                      grid[t[0]][t[1] + 1] = 2;
                      list.add(new int[]{t[0], t[1] + 1});
                  }
              }
          }
          return freshCount == 0 ? res : -1;
      }
      

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• the most votes
  • BFS Runtime: 3 ms, faster than 82.65%, Memory Usage: 38.8 MB, less than 94.48% of Java online submissions

    // O(r * c)time
    // O(Max(len(queue)))space
    public int orangesRotting(int[][] grid) {
        if(grid == null || grid.length == 0) return 0;
        int rows = grid.length;
        int cols = grid[0].length;
        Queue<int[]> queue = new LinkedList<>();
        int count_fresh = 0;
        //Put the position of all rotten oranges in queue
        //count the number of fresh oranges
        for(int i = 0 ; i < rows ; i++) {
            for(int j = 0 ; j < cols ; j++) {
                if(grid[i][j] == 2) {
                    queue.offer(new int[]{i , j});
                }
                else if(grid[i][j] == 1) {
                    count_fresh++;
                }
            }
        }
        //if count of fresh oranges is zero --> return 0
        if(count_fresh == 0) return 0;
        int count = 0;
        int[][] dirs = {{1,0},{-1,0},{0,1},{0,-1}};
        //bfs starting from initially rotten oranges
        while(!queue.isEmpty()) {
            ++count;
            int size = queue.size();
            for(int i = 0 ; i < size ; i++) {
                int[] point = queue.poll();
                for(int dir[] : dirs) {
                    int x = point[0] + dir[0];
                    int y = point[1] + dir[1];
                    //if x or y is out of bound
                    //or the orange at (x , y) is already rotten
                    //or the cell at (x , y) is empty
                        //we do nothing
                    if(x < 0 || y < 0 || x >= rows || y >= cols || grid[x][y] == 0 || grid[x][y] == 2) continue;
                    //mark the orange at (x , y) as rotten
                    grid[x][y] = 2;
                    //put the new rotten orange at (x , y) in queue
                    queue.offer(new int[]{x , y});
                    //decrease the count of fresh oranges by 1
                    count_fresh--;
                }
            }
        }
        return count_fresh == 0 ? count-1 : -1;
    }
    

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