994. Rotting Oranges (Python)

Related Topic

Breadth-First-Search.

Description

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.

Sample I/O

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

Methodology

1. Add all fresh oranges to fresh_set and append all rotten oranges to rotten_queue.
2. Use BFS to find all fresh oranges that adjacent to the current rotten orange, turn these fresh oranges to rotten and remove these from fresh_set. In the meantime, track the steps of turning.
3. After we finish the turning, if there is still a fresh orange in fresh_set, return -1 otherwist return the step.

Code (BFS)

def orangesRotting(self, grid: List[List[int]]) -> int:
        row, col = len(grid), len(grid[0])
        dirs = [(-1,0),(0,1),(1,0),(0,-1)]
        fresh_set=set()
        rotten = collections.deque()
        step = 0
        for x in range(row):
            for y in range(col):
                if grid[x][y]==1:
                    fresh_set.add((x,y))
                elif grid[x][y]==2:
                    rotten.append([x,y,step])
        while rotten:
            x,y,step = rotten.popleft()
            for dx, dy in dirs:
                if 0<=x+dx<row and 0<=y+dy<col and grid[x+dx][y+dy] == 1:
                    grid[x+dx][y+dy]=2
                    fresh_set.remove((x+dx,y+dy))
                    rotten.append([x+dx,y+dy,step+1])
        return step if not fresh_set else -1

BigO

We traversal all elements of 2D list once so total time complexity is O(mn) where m is size of row and n is size of columns