Rotting Oranges

Posted: 13 Sep, 2020
Difficulty: Moderate


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You have been given a grid containing some oranges. Each cell of this grid has one of the three integers values:

  • Value 0 - representing an empty cell.
  • Value 1 - representing a fresh orange.
  • Value 2 - representing a rotten orange.
  • Every second, any fresh orange that is adjacent(4-directionally) to a rotten orange becomes rotten.

    Your task is to find out the minimum time after which no cell has a fresh orange. If it's impossible to rot all the fresh oranges then print -1.

    1. The grid has 0-based indexing.
    2. A rotten orange can affect the adjacent oranges 4 directionally i.e. Up, Down, Left, Right.
    Input Format:
    The first line of input contains two single space-separated integers 'N' and 'M' representing the number of rows and columns of the grid respectively.
    The next 'N' lines contain 'M' single space-separated integers each representing the rows of the grid.
    Output Format:
    The only line of output contains a single integer i.e. The minimum time after which no cell has a fresh orange. 
    If it's impossible to rot all oranges, print -1.
    You are not required to print the expected output, it has already been taken care of. Just implement the function.
    1 <= N <= 500
    1 <= M <= 500
    0 <= grid[i][j] <= 2
    Time Limit: 1 sec
    Approach 1

    The idea is very simple and naive. We will process the rotten oranges second by second. Each second, we rot all the fresh oranges that are adjacent to the already rotten oranges. The time by which there are no rotten oranges left to process will be our minimum time.


    In the first traversal of the grid, we will process all the cells with value 2 (rotten oranges). We will also mark their adjacent cells as rotten for the next traversal. Now, we can’t mark them by assigning the same value i.e. 2 because then we won’t able to differentiate between the current processing cells and the cells which are going to be processed in the next traversal. So, we will mark them as value 3. More formally, we will be marking the adjacent cells as ‘CURR_ROTTEN’ + 1 in each traversal of the grid.


    Here is the complete algorithm.

    • Initialize ‘TIME’ to 0 and ‘CURR_ROTTEN’ to 2.
    • Loop until there is no rotten orange left to process.
      • Initialize ‘NOT_FOUND’ to true.
      • We will run a nested loop and traverse the grid.
        • If the element of the grid is equal to ‘CURR_ROTTEN’ then we have to process this rotten orange.
        • So we will assign the adjacent elements to (‘CURR_ROTTEN’ + 1), if the adjacent orange is fresh (value is 1), and ‘NOT_FOUND’ to false.
      • If ‘NOT_FOUND’ is true, break.
      • Else, increment ‘TIME’ by 1.
      • Increment ‘CURR_ROTTEN’ by 1.
    • At last, we will traverse the grid and check if there is any fresh orange left i.e. a cell with value 1. If found, return -1.
    • Else return time elapsed i.e. maximum of ‘TIME’ - 1 and 0.
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